Electrostatics 11

Grade 11 © GC Shiba
Electrostatics Grade 11





GC Shiba



























Electricity and Magnetism ( 27 teaching hours)




• Electric charge, conductors and insulators, Charging by induction,
Coulomb's law - Force between two-point charges, Force between multiple
electric charges (3 hrs)


• Electric fields, Calculation of electric field due to point charges, Field lines,

Gauss law, Electric flux, Gauss law and its application, field of a charged
sphere, line charge, charge plane conductor (3 hrs)


• Electric potential, potential difference, potential due to a point charge,
equipotential lines and surfaces, potential gradient, potential energy, electron
volt (4 hrs)


• Capacitance and dielectrics: Capacitance and capacitor, parallel plate
capacitor, combination of capacitors, energy of charged capacitor, effect of

a dielectric, polarization and displacement (7 hrs)


• DC Circuits: Electric Currents; Drift velocity and its relation with current,

Ohm’s law; Electrical Resistance; Resistivity; Conductivity, Current-

voltage relations; Ohmic and Non-Ohmic resistance, Resistances in series

and parallel, Potential divider, Electromotive force of a source, internal

resistance, Work and power in electrical circuits (10 hrs)


2 Electrostatics

Grade 11 © GC Shiba
Contents



Electricity: ........................................................................................................................................................ 4

Electrostatic induction: ................................................................................................................................... 5

Action of points: ................................................................................................................................................ 6

Coulomb's law of electrostatics: ..................................................................................................................... 7

Electric field intensity: ..................................................................................................................................... 9

Electrical potential ......................................................................................................................................... 10

Electric potential energy................................................................................................................................ 11

Electric potential difference: ......................................................................................................................... 13
Electric intensity at any point on the axis of dipole: ................................................................................... 15

Electric potential gradient: ........................................................................................................................... 17

Relation between Electric potential gradient and Electric intensity: ................................................................... 17
Equipotential surface: .................................................................................................................................... 18

Electron volt ................................................................................................................................................... 18

Electric field .................................................................................................................................................... 18

.................................................................................................................................................................................... 19
Properties of electric lines of force: ......................................................................................................................... 19

Electric flux: ................................................................................................................................................... 21

Gauss Theorem: ............................................................................................................................................. 21

Applications of Gauss Theorem: .................................................................................................................. 22
1) Electric intensity inside ..................................................................................................................................... 23
2) Electric intensity outside .................................................................................................................................. 23

3) Electric intensity on the surface ....................................................................................................................... 24
4) Electric intensity due to linear distribution of charge ................................................................................... 25

Electric intensity due to charged plane sheet ......................................................................................................... 26
Electric intensity due to charged plane conductor ................................................................................................. 27
Electrostatic shielding ........................................................................................................................................... 27
Zero potential: ....................................................................................................................................................... 27

Short Questions ( 2 Marks) .................................................................................................................. 28

Long Answer Questions (4 Marks) .................................................................................................... 29

Numerical Problems (3 Marks) .......................................................................................................... 30







3 Electrostatics

Grade 11 © GC Shiba

Electricity: The branch of physics which deals with charges either at rest or in
motion is called electricity. There are two branches of electricity.


1. Electrostatics: The branch of electricity that deals with a charge at rest.

2. Current electricity: The branch of electricity that deals with a charge in
motion.
Electric charge: It is defined as the product of electric current and time for which
this current persists.

i.e., q = I.t
☻ Electric charge exists either positive or negative.
☻ Like charges repel and unlike charges attract.

☻ It is a scalar quantity.
☻ Electric charge is additive. The total electric charge of a body is the algebraic
sum of all charges on it.
☻ It is quantized. It means charge is always an integral multiple of basic charge

e (charge of an electron).
-19
i.e., Q = ± n e ; where n is an integer and e = -1.6 × 10 C.
☻ Total charge in an isolated system remains constant. This is called the
principle of conservation of charge.
☻ The electric charge on a body doesn't depend on its speed.
☻ SI unit of charge is coulomb (ampere sec).

9
☻ CGS unit of charge is stat coulomb ( 1 C = 3 × 10 stat coulomb ).



Conductors and insulators:


i) Conductor: A material which allows charge to move in it due to free
electrons is called conductor. E.g., silver, copper, carbon in the form of
charcoal and graphite, coal, acid, alkali, salt , earth human body etc.

ii) Insulators: A material which does not allow charges to move in it is called
an insulator. E.g., non-metals, glass, rubber, plastics, paper, dry wood,
mica, porcelain, dry air, ebonite etc.
iii) Semiconductors: A material which behaves as conductors under certain

physical conditions is called semiconductors. E.g., silicon, germanium,
carbon etc.



Electrification: The process of charging a body positively or negatively by
transfer of electrons in the outermost orbits from one body to the other.






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Charging a body: The process in which a body gains or loses charge particles

(electrons). There are three methods of charging a body.

1 By friction: The process in which two bodies are rubbed against each other

due to which there is a transfer of electrons between two bodies and one
becomes positively charged while another becomes negatively charged.
2 By conduction: The process of charging a body by touching with a charged

body in which the same nature of charge is produced on the conductor.



Conductor Positively charged body








Insulated stand


3 By induction: The process of charging a body without touching the charged
body.


Induced charge
Inducing charge







Insulated stand

Electrostatic induction:


When a charged body is brought close to the conductor, the opposite charge

is developed at the near end of the conductor while a similar charge is developed
at the far end of the conductor temporarily so long as the charged body is close to
the conductor. This phenomenon is known as electrostatic induction.


Developed charge in a conductor is called induced charge and charge of the
charged body is called inducing charge.


Inducing charge and induced charge of each nature are equal in magnitude.


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Distribution of charge over the surface of conductor

☻ When a body is charged, the charge spreads all over the surface of the body

which may or may not be uniform.
☻ The distribution of charge depends upon the shape of the conductor.
☻ The greater the curvature at any point, the greater will be the accumulation of
charge.

Types of charge distribution:
1. Linear charge density (λ ): The charge per unit length of a body is called
linear charge density. i.e., λ =q/l

The charge distributed on a very thin straight rod.
2. Surface charge density (σ ): The charge per unit area of a body is called
surface charge density. i.e., σ =q/A
The charge distributed uniformly on a thin disc.

3. Volume charge density (ρ ): The charge per unit volume is called volume
charge density. i.e., ρ = q/V

The charge distributed inside a spherical non-conducting body.




Action of points:


The leakage of charge from sharp points is called action of point.
☻ The distribution of charge over the surface of a conductor is uniform

only when the surface is uniform shaped.
☻ The density of charge at any part of
conductor is inversely proportional to the

radius of curvature of the surface of that part.

i.e., (σ = ⟹ ∝ )

☻ It is greater on those parts of the surface

which have the greater curvature (least radius of curvature).
☻ Hence, density of charge at the pointed parts of a conductor is very large.
☻ E.g., The air particles and dust particles coming in contact gain charge
from the sharp points by conduction.

Uses of Action of points:
☻ The lightning conductor are made pointed upward in order to have a safe

discharge of the electricity generated in the atmosphere.
☻ The electric poles supporting the transmission lines at high voltage are
fitted with a sharp pointed conductor at the top. If there is any leakage of
charge from the line to the poles, it is discharged into the atmosphere

through the pointed ends.
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☻ Sharp edges or points are strictly avoided in electrical machines.


Coulomb's law of electrostatics:



The force acting between two charges is


➢ directly proportional to the product of charges
➢ inversely proportional to the square of distance between the charge.


Q1 Q2

r

Let us consider two charges Q1 and Q2 are at a distance r. The force
acting between two charges is F.

According to Coulomb's law,
The force acting between two charges is
i. directly proportional to the product of charges

i.e., ∝ ..................... ( i )
1 2
ii. inversely proportional to the square of distance between the charge.
1
i.e., ∝ ..................... (ii)
2
Combining equations (i) and (ii), We get


1 2
∝
2


= 1 2 ; where k is proportionality constant.
2


The value of k depends upon unit chosen and medium on which charges are
kept.

For free space in cgs unit, k = 1


1 2
∴ =
2
For free space in SI unit, k is replaced by new constant 1 Where ∈ is
0
4 ∈ 0

2
-1
-1
-2
permittivity of the free space having value 8.85 × 10 −12 C N m (Fm )

9
-2
2
and the value of 1 = 9 × 10 Nm C .
4 ∈ 0
7 Electrostatics

Grade 11 © GC Shiba



1 2
∴ =
4 ∈ 2
0
For medium in SI unit, k is replaced by new constant 1 Where ∈ is
4 ∈
permittivity of the medium.

1 2
∴ =
4 ∈ 2

Permittivity: It is the ability to permit the electric force by medium or free
space.



Dielectric constant/ relative permittivity (k):

Dielectric constant of a medium is defined as the ratio of force acting
between any two charges in free space at a certain distance to the force
acting between two same charges in a medium at same distance.



Mathematically, k (∈ ) =




1 2
4 ∈ 2
= 0

1 2
4 ∈ 2

∈
∴ =
∈ 0
Note: The dielectric constant depends on other factors like temperature,
frequency of voltage applied, moisture etc. It increases with increase in
temperature in solid and decreases with increase in temperature in gas and

liquid.

1 Coulomb: Two similar charges are said to be 1 coulomb if they exert the
9
force of 9 × 10 N on each other at a distance of 1 meter in free space.

Limitations of Coulomb's law:
☻ It holds good for stationary point charges only.

☻ it is not universal as it depends on properties of intervening medium.



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Electric field intensity:



Electrical intensity at any point in the electric field is defined as the force
experienced by unit positive charge (+1) placed there.



P

+Q
r (+1)



let us consider a point P in the electric field of +Q charge at a distance r. We

want to find electric intensity at point P. Put +1 charge (test charge) at point P.

The electric intensity at a point P between +Q and +1 according to Coulomb's

1 2
formula is = Where is permittivity of the free space.
0
4 2
0
(+ )(+1)
=
4 2
0

=
4 2
0

By definition, this force is electric intensity (E).

∴ =
4 2
0

This is the required expression for electric intensity. It is a vector quantity. Its
direction is radially away from the +Q charge and radially towards the -Q charge.


Force between Multiple electric charges:
Electric force at a point due to the number of charges is vector sum of the
forces due to individual charges.


P
Fn
F1

F3 Qn
F2

Q1
Q3

Q2


E = E1 + E2+ E3+ ... ... ... + En ;Where E stands for force.
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Electric potential: It is measured in terms of work done for which a reference

point is considered as infinity where potential is zero. (from Coulomb's law we
1
know that ∝ i.e., when → ∞, → 0).
2


Electric potential = [ ]
ℎ


or, = [ ]


Its SI unit is Volt (joule per coulomb) and CGS unit is stat volt.


1 volt = 1/300 stat volt


1 volt: Electric potential at a point in the electric field is said to be 1 V if 1 joule
work is done in a moving charge of 1 coulomb from infinity to that point.


☻ Define electric potential and derive an expression for it due to point charge.


Electrical potential at any point in the electric field is defined as the amount

of work done in taking unit positive charge (+1) from infinity to the point under
consideration.


P dr
⃗⃗

+Q
r (+1)



let us consider a point P in the electric field of +Q charge at a distance r. We

want to find electric potential at point P. Put +1 charge at point P.

The electric force acting at a point P between +Q and +1 according to Coulomb's

formula is


= Where is permittivity of the free space.
0
4 2
0
Now +1 charge is displaced by very small distance dr towards +Q charge then


⃗ ⃗⃗⃗⃗⃗
work done (dw) = .

= F dr cos180 0


= - F dr



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= −
4 2
0

Now, total work done in displacing +1 charge from infinity to point P is integral
of dw between limit ∞ to r.



∴ W = ∫
∞

= ∫ − 2
0
∞ 4
1
= − ∫
4 0 ∞ 2



= − ∫ −2
4 0 ∞


−2+1
= − [ ]
4 0 −2 + 1 ∞


1 1
= [ − ]
4 0 ∞


1
= [ ]
4 0


By definition, this work done is electric potential (V).

∴ =
4
0

This is the required expression for electric potential.





Electric potential energy: Electrical potential energy at any point in the

electric field is defined as the amount of work done in taking a given charge from

infinity to the point under consideration.




P dr
⃗⃗
+Q
r (+q)


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let us consider a point P in the electric field of +Q charge at a distance r. We
want to find electric potential energy at point P. Put +q charge at point P.


The electric force acting at a point P according to Coulomb's formula is



= Where is permittivity of the free space.
0
4 2
0
Now +q charge is displaced by very small distance dr towards +Q charge then


⃗ ⃗⃗⃗⃗⃗
work done (dw) = .

= F dr cos180 0


= - F dr



= −
4 2
0
Now, total work done in displacing +q charge from infinity to point P is integral
of dw between limit ∞ to r.



∴ W = ∫
∞


= ∫ − 2

0
∞ 4
1
= − ∫
4 0 ∞ 2



= − ∫ −2
4 0 ∞


−2+1
= − [ ]
4 0 −2 + 1 ∞


1 1
= [ − ]
4 0 ∞

1
= [ ]
4 0


By definition, this work done is electric potential energy (U).
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∴ =
4
0

This is the required expression for electric potential energy.


☻ Use Coulomb's law to define potential difference between two
points near a static charge. Also derive expression for it.




Electric potential difference:


Electrical potential difference between two points in the electric field is defined as

the amount of work done in taking unit positive charge from one point to another
point.



r2 A P B
+Q
r1 (+1)

r

Consider two points A and B at distances r1 and r2 from +Q charge. We want
to find p.d. between points A & B. Let us take another point P at a distance f of +Q

charge on the same line. Put +1 charge at point P.

Electric force at point P according to Coulomb's law is



= Where is permittivity of the free space.
0
4 2
0
Now +1 charge is displaced by very small distance dr towards +Q charge then


⃗ ⃗⃗⃗⃗⃗
work done (dw) = .

= F dr cos180 0


= - F dr



= −
4 2
0
Now, total work done in displacing +1 charge from B to A is integral of dw

between limit r2 to r1.


1
∴ W = ∫
2

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1
= ∫ − 2
0
2 4

1 1
= − ∫
4 0 2 2


1
= − ∫ −2
4 0 2



−2+1 1
= − [ ]
4 0 −2 + 1 2


1 1
= − [ − ]
4 0 1 2


− 1
2
= − [ ]
4 0 . 2
1
By definition, this work done is electric potential difference (V).
− 1
2
∴ = − [ ]
4 0 . 2
1

This is the required expression for electric potential difference between
points A and B.


Electric potential at a point due to number of charges:


Electric potential at a point due to the number of charges is the algebraic sum
of electric potentials due to individual charges.





P

rn
r1

r3 Qn
r2

Q1
Q3

Q2


V = V1 + V2+ V3+ ... ... ... + Vn



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Electric intensity at any point on the axis of dipole:



Dipole: Two equal and opposite charge at a distance is called a dipole.


Dipole moment (p): It is the product of one or two equal charges at dipole and
distance between them. i.e., p = q x 2l.



A O B r P


+q -q
L L


Let us consider a dipole of dipole moment P. Here +q and -q charges are at a
distance 2L and O is the Centre of dipole. Let OP = r and from figure, AP = (r +

L) and BP = (r – L)


Electric intensity at point P due to charge at point A is


⃗⃗⃗⃗⃗⃗
=

4 ( + ) 2
0
Electric intensity at point P due to charge at point B is


⃗⃗⃗⃗⃗⃗
= 4 ( − ) 2

0
Electric intensity at point P due to dipole is = −


= ( 1 − 1 )
4 0 ( − ) 2 ( + ) 2


2
( + ) − ( − ) 2
= ( )
2
4 0 ( − ) ( + ) 2
4
= ( )
2
2 2
4 0 ( − )
2( × 2 )
=
2
2 2
4 ( − )
0
2
∴ = (∵ = × 2 )
2
2 2
4 ( − )
0
This is the required expression for intensity at point P due to dipole.



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Electric intensity at any point on the broadside on position or bisector of

dipole:
P
Let us consider a dipole of dipole moment P.

Here +q and -q charges are at a distance 2L and O is
the Centre of dipole. Let OP = r and from figure, r

2
2
2
2
= √ + and = √ + .
A B
Electric intensity at point P due to charge at point A +q O -q
is

⃗⃗⃗⃗⃗⃗
=

2
2
4 ( + )
0
Electric intensity at point P due to charge at point B is

⃗⃗⃗⃗⃗⃗
=

2
2
4 ( + )
0
Electric intensity at point P due to dipole is

= √ 2 + 2 + 2 2



, = ∠ = ∠


, =


= √1 + 1 + 2 2


= √2 + 2 2


= √2 + 2 2


= √2 √1 + 2 2

2
= 2 (∵ 2 = 2 − 1)


= 2 × ×
2
2
4 ( + ) √( + )
2
2
0

∴ = (∵ = × 2 )
2 3/2
2
4 ( + )
0
This is the required expression for intensity at point P due to dipole.
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☻ Define electric field intensity and potential gradient and


establish a relation between them.



Electric potential gradient:



Electric potential gradient in electric fields is defined as the rate of change of
-1
electric potential with distance. Its SI unit is vm .

let ∆ be the change in electric potential in distance ∆ then


Electric potential gradient = ∆
∆
∆
= lim ( )
∆ →0 ∆


=



Relation between Electric potential gradient and Electric intensity:


A B


+Q
r
r + Δr

r
Consider two points A and B in the electric field of +Q charge at distances r
and r + Δr respectively. Type equation here.

Let electric potential at point A is =V

& electric potential at point B is = + Δ

P.d. between A and B, = − = − ( + Δ )


Work done to move +1 charge from A to B = −Δ


Force on +1 charge × Δr = −Δ


or, E × Δr = −Δ


Δ
or, = −
Δ

or, = −





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Thus, Electric intensity is equal to negative of electric potential gradient. Direction

of E is along which electric potential gradient decreases.



Equipotential surface:


Equipotential surface is such a surface on
which electric potential is the same everywhere. It

means the potential difference between any two
points on an equipotential surface is zero. We know
that p.d. is work done and hence there is no work is

done in moving the charge from one point to another on
an equipotential surface. E


❖ Electric intensity is always normal to the equipotential surface.
❖ Surface of a charge sphere is always an equipotential surface.
❖ In the region of high electric intensity, equipotential surface is closer while
in the region of low electric intensity, equipotential surface is farer.

❖ Two equipotential surfaces are never intersecting.





Electron volt: The kinetic energy gained by an electron when accelerated
through a potential difference of one volt.


i.e., 1 eV = electric work done to accelerate the electron through a p.d. of 1 volt


= K.E. gained by electron


= loss in P.E.


∴ 1 = 1.6 × 10 −19 × 1

= 1.6 × 10 −19


= 1.6 × 10 −19






Electric field: The region around a charged body up to which it exerts electric
force on any other charge is called electric field.


Electric lines of force: It is defined as the imaginary path along which a unit
positive charge would move if it were to move.


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Electric lines of force in the electric field are imaginary continuous curves

such that tangent to any point of curve gives the direction of the electric field at
that point.































Properties of electric lines of force:


☻ The electric lines of force originate normally from positive charge and
terminate normally at negative charge.


























☻ The electric lines of force do not pass through a conductor.

☻ It is not affected by earth's magnetic field.
☻ These are not closed curves like magnetic lines of force.









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☻ The tangent at any point on the lines of force gives the direction of the

resultant electric field intensity at that point.

E



E











Fig. Electric field along tangent at line of force


☻ They are like stretched elastic strings which tend to contract lengthwise and
repel sidewise.



























Fig. Lines of force for unlike charges Fig. Lines of force for like charges




☻ Two electric lines of forces never intersect. If
they intersect each other, there must be two
directions of electric field intensity at the point
of intersection which is impossible.













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Electric flux: Electric flux is the
number of electric lines of force passing

normally through a certain area. It is
2
-1
denoted by . Its SI unit is Nm C .



Electric flux density: Electric flux density is the number of electric lines of force

passing normally through unit area. It is denoted by E.
Electric flux density is measured by the electric intensity since the lines of force
are closer to each other in the high electric intensity region and away from each

other in the low electric field.

Mathematically, =



∴ = .




☻ State and explain Gauss theorem and use it to find the electric

field due to a charged sphere at a point

i. Outside the sphere and

ii. Inside the sphere




Gauss Theorem:


Statement: The total electric flux passing through the closed surface of

1
arbitrary shape is equal to times the enclosed charge where ∈ is the
∈ 0 0
permittivity of a free space.
ℎ
∴ =
∈ 0





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Let +Q charge enclosed by an arbitrary shape surface is shown in figure.



























Total electric flux through this surface is


ℎ
=
∈ 0

∴ =
∈ 0
Proof for spherical surface:
Consider a +Q charge at the center of a sphere of radius r. The electric
intensity at the surface of sphere is


=
4 2
0
1
= Where = 4 = ℎ
2

0

or, E . A =
∈ 0

∴ =
∈ 0







Applications of Gauss Theorem:


In actual practice, we can never have a point charge, rather we do have
charge bodies having different shapes. Coulomb law cannot be applied directly to

calculate the electric field having different shapes of charge bodies. In such
situations, Gauss theorem is applicable.


22 Electrostatics

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1) Electric intensity inside a charged conducting solid sphere
Consider a charge conducting sphere of radius R

having a charge +Q. We are interested to find the electric
intensity at point P inside a charged sphere. Let us draw a
concentric sphere of radius r through point P. This sphere is P

called a Gaussian sphere.
By symmetry, electric intensity would be the same at all
points on the surface of a Gaussian sphere. Electric flux from

the charged sphere is radially outward as shown in the
figure.
Now, electric flux passing through the gaussian surface is

ℎ
=
∈ 0

=E . A
or, E . A = 0 ( since no charge is enclosed by
gaussian surface)

E = 0
Hence, there is no electric field inside the charged sphere.










2) Electric intensity outside a charged conducting sphere
Consider a charge conducting sphere of radius R

having a charge +Q. We are interested to find the electric
intensity at point P outside a charged sphere. Let us draw

a concentric sphere of radius r through point P. This
sphere is called a Gaussian sphere.
By symmetry, electric intensity would be the same at all
points on the surface of a Gaussian sphere. Electric flux

from the charged sphere is radially outward as shown in
the figure.
Now, electric flux passing through the gaussian surface is

ℎ
=
∈ 0

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=E . A

or, E . A =
0
=
0

=
4 2
0


Hence, charged spheres behave as whole charges are concentrated at its
center.


3) Electric intensity on the surface of a charged conducting

sphere

Consider a charge conducting sphere of radius R
having a charge +Q. We are interested to find the
electric intensity at point P on the surface of a charged

sphere. Let us draw a concentric sphere of radius r
through point P. This sphere is called a Gaussian sphere. r = R P
By symmetry, electric intensity would be the same at all

points on the surface of a Gaussian sphere. Electric flux
from the charged sphere is radially outward as shown in
the figure.

Now, electric flux passing through the gaussian surface
is
ℎ
=
∈ 0

=E . A


or, E . A =
0
=
0

=
4 2
0

Hence, the electric field just outside the surface.






24 Electrostatics

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☻ State and explain Gauss law in electrostatics. Use it to find the

electric field intensity due to a line charge.



4) Electric intensity due to linear distribution of charge (a thin

straight charged wire or rod)

Let us consider a thin long straight wire XY having X
linear charge density (λ = charge/length). Let p be
a point at a perpendicular distance r from the line

charge where electric intensity is to be determined.
The direction of electric flux is radially outward. A
cylindrical gaussian surface of radius r, length l and
coaxial with the line charge is drawn as shown in

the figure.


The total electric flux passing through Gaussian
cylinder (hypothetical) according to Gauss theorem is


ℎ Y
=
∈ 0

=E × A (where E is electric intensity on the
surface of Gaussian cylinder and A

is surface area of Gaussian cylinder)

or, E . A =
0

=
2 0



=
2 0


Hence, the electric field intensity of a charged wire is inversely

1
proportional to the distance from the wire ( = ).







25 Electrostatics

Grade 11 © GC Shiba
☻ What is electric flux? State and explain Gauss law in electrostatics.

Use it to find electric field intensity due to infinite plane sheet of

charge.



Electric intensity due to charged plane sheet


Let us consider a positively

charged plane sheet having surface
charge density σ .( σ =
charge/surface area). We are
interested to find the electric

intensity at point P outside the
charged plane sheet at a distance r.

A cylindrical Gaussian surface PP'
of cross-sectional area A through
P is drawn as shown in figure.


The total electric flux passing through the two flat surfaces at the ends of PP' is

= 2 .


Since net charge enclosed by the area A is σ A, so by Gauss's theorem,

ℎ
=
∈ 0



, 2 . =
∈ 0


=
2 0


Clearly, E is independent of r, remains the same and does not change
with distance from the conductor. The field lines remain everywhere

straight, parallel and equally spaced.







26 Electrostatics

Grade 11 © GC Shiba
☻ State Gauss law of electrostatics and use it to find the electric


field intensity due to a plane charge conductor.



Electric intensity due to charged plane conductor


Let us consider a positively charged

plane conductor having surface charge
density σ .( σ = charge/surface area). We
are interested to find the electric intensity at
point P outside the charged plane conductor

at a distance r. A cylindrical Gaussian
surface of cross-sectional area A through P
is drawn as shown in figure.

The total electric flux passing through the
Gaussian surface is = .
Since net charge enclosed by the area A is σ A, so by Gauss's theorem,
ℎ
=
∈ 0



, . =
∈ 0


=
0
This is the required expression for electric intensity due to the charged plane

conductor.

Electrostatic shielding: The electric field inside a conductor or cavity is always

zero which is not influenced by external electric field whatever be the charge and


field configuration outside (the region where no electric shock is experienced due
to zero potential). This is known as electrostatic shielding. E.g., a car has a hollow


metallic body and we remain safe inside it due to electrostatic shielding during
thunderstorm and lightning.



Zero potential: If a body can either supply or absorb any amount of charge without

charge in electric potential, it is said to be at zero potential. Electric potential of

earth is said to be zero potential. It is reference potential.


27 Electrostatics

Grade 11 © GC Shiba
Short Questions ( 2 Marks)





1 Why metallic chain is used in the tanker transporting highly inflammable
liquid?

2 Is it possible that electric potential at a point is zero but not electric field?

3 How can a body be charged with positive electricity by the method of
induction? Explain.

4 What is the magnitude of an electric field which will balance the weight of
an electron on the surface of earth?

5 Can two electric lines of force ever intersect? Explain.

6 Different charges q 1, q 2, ... are placed at r 1, r 2, ... respectively in an
electrostatic field. Write expressions for the total electric potential and

intensity at R in proper notations. r1, r2, ... and R refer to the distances from

the origin of the field (space).
7 Define and write the unit of potential gradient in an electrostatic field.

8 Explain how a pith ball can be electrified with positive or negative charge.

9 What is potential gradient? How is it related with the electric field intensity?
10 Can a charged body attract an uncharged body?

11 A man inside an insulated metallic cage does not receive shock when the

cage is highly charged, why?
12 Why does body get electrified when they are rubbed together for a while?

13 What is electrostatic shielding?
14 Two charged conductors are touched mutually and then separated. What

will be the charge on them?

15 How can a neutral body be charged with negative electricity by induction?
16 Two electric lines of force never intersect each other. why?

17 The tyres of airplane are made from a special rubber which is reasonably

good conductor of electricity. What would be the reason behind it?
18 More charge can be stored on a metal if it is highly polished that when its

surface is rough. Explain.
19 Why are the four-footed animals posed to more threat during close by

lightning strikes than the two footed humans?

20 Why is it dangerous to take shelter under a tall tree during lightening?
21 Sharp projections are avoided in machines, why?



28 Electrostatics

Grade 11 © GC Shiba

22 What do you mean by quantization of charge?
23 What do you mean by charging by conductor and charging by induction?

24 Define one electron volt.

25 If the electric field is zero throughout a certain region of space, is the
potential also zero in the region or not? Explain.

26 A comb run through one's dry hair attracts bits of paper, why?
-1
-1
27 Prove 1 Vm = 1 NC .
28 Why do sharp edges are strictly avoided in an electrical machine?

29 What is meant by relative permittivity? What is its minimum value?
30 Explain the phenomenon of action point in a charged conical shape.

31 A charged conical conductor loses its charge earlier than a similarly charged

sphere, Why?

32 Why are sharp edges or points avoided in electrical machines?


Long Answer Questions (4 Marks)




☻ Define electric field intensity. Write down its units. Obtain an expression for

electric field due to an electric dipole at a point equidistance from each
charge.

☻ State Gauss law of electrostatics and use it to find the electric field intensity

due to a plane charge conductor.
☻ Define electric potential and derive an expression for it due to point charge.

☻ State and explain Gauss law of electrostatics. Apply it to obtain an

expression for electric field of a linearly charged body.
☻ Are electrical potential and electric field vector quantities? Justify with their

definitions. Also obtain an expression for electrical potential at a point due

to a point charge.
☻ Use Coulomb's law to define potential difference between two points near

a static charge. Also derive expression for it.
☻ State and explain Gauss law of electrostatics with one of its applications.

☻ Define electric field intensity and potential gradient and establish a relation

between them.
☻ What is electrostatic induction? Explain with necessary diagram, a method

of charging a body positively by induction.



29 Electrostatics

Grade 11 © GC Shiba

☻ State and explain Gauss theorem and use it to find the electric field due to
a charged sphere at a point

a. Outside the sphere and

b. Inside the sphere
☻ What is electric flux? State and explain Gauss law in electrostatics. Use it to

find electric field intensity due to infinite plane sheet of charge.
☻ Define electric field intensity and dielectric constant of a medium. Also

derive a relation between the electric field with the potential gradient.

☻ What is the physical difference between electric potential and electric field
intensity? Derive an expression for the potential differences between two

points in the electrostatic field.

☻ Define electric potential. Derive an expression for the electric potential due
to a point charge at a distance r from it.

☻ State and explain Gauss law in electrostatics. Use it to find the electric field

intensity due to a line charge.
☻ State and explain Gauss theorem in electrostatics and use it to find the

electric field intensity due to a hollow charged spherical conductor.

☻ Define potential and electric field at a point in an electrostatic field. Derive
a relation between the electric potential and the electric field strength at a

point.



Numerical Problems (3 Marks)




1 Find the electric intensity at corner C of equilateral triangle of side 1 m if
-6
-6
2 × 10 C and -3 × 10 C are at corner of A and B of triangle.
4
4
Ans: 1.8 × 10 N/C along AC and 2.7 × 10 N/C along BC
2 Find electric potential at corner C of equilateral triangle of side 1 m if
-6
-6
2 × 10 C and -3 × 10 C are at corners A and B of a triangle.
Ans:

3 Two large parallel metal plates carry opposite charges. They are separated
by 10 cm and p.d. of 500 volts is applied on them.

a. What is the magnitude of electric field strength between them?
−9
b. Compute the work done by this field on a charge of 2 × 10 as it
moves from higher to the lower potential.

-6
Ans: 10 J
30 Electrostatics

Grade 11 © GC Shiba

4 Two large parallel metal plates carry opposite charges. They are separated
by 0.20 m and the p.d. between them is 500 V. What is the magnitude of

electric field, if it is uniform, in the region between them?

Ans: 2500 V/m
5 What distance must an electron move in a uniform potential gradient

-18
-1
200 V cm in order to gain kinetic energy 3.2 × 10 ?
Ans: 0.001 m
-19
-31
6 An electron of mass 9.1 × 10 kg and charge 1.6 × 10 C is situated in a
4
-1
uniform electric field of intensity 1.2 × 10 Vm . Find the time it takes to
travel 1 cm from rest.
-9
Ans: 3.1 × 10 Sec
7 A solid sphere of radius 1 cm is carrying a charge of 2 C. Find the electric
field intensity at the centre, on its surface and at a point 2 cm from the

centre of the charged sphere.

12
13
Ans: 1.8 × 10 N/C, 4.5 × 10 N/C
8 An alpha particle is a nucleus of doubly ionized helium. It has mass of 6.68
-19
-27
× 10 kg and charge of 3.2 × 10 C. Compare the force of electrostatic
repulsion between the two alpha particles with the force of gravitational
attraction between them.

35
Ans: 3.1 × 10 N
-19
9 An electron of charge 1.6 × 10 C is situated in a uniform electric field
intensity 120,000 V/m. Find the force on it and its acceleration.
16
-14
Ans: 1.92 × 10 N, 2.1 × 10 m/s 2


10 Calculate the value of two equal charges if they repel one another with a

force on 0.1 N when situated 50 cm apart in vacuum. What would be the
distance between them if they are placed in an insulating medium of

dielectric constant 10 ?
-6
Ans: 1.67 × 10 C, 0.16 m
6
-6
11 Two charges +1 × 10 and -4 × 10 C are separated by a distance of 2 m.
Determine the position of null point.
Ans: 2 m from small charge

12 Two horizontal parallel plates, 20 mm apart, have a potential difference of

1000 V between them the upper plate being at a positive potential. A
-15
negatively charged oil drop, mass 4.8 × 10 Kg, is situated between the
31 Electrostatics

Grade 11 © GC Shiba

plates. Calculate the number of electrons on the drop if it is stationary in
the air, neglecting the density of air.

Ans: 6
-6
13 A hollow spherical conductor of radius 12 cm is charged to 6 × 10 C. Find
the electric field strength at the surface of sphere, inside the sphere at 8

cm and at a distance 15 cm from the sphere.
6
5
Ans: 3.75 × 10 N/C, 0, 7.41 × 10 N/C
2
14 Two horizontal parallel plates each of area 500 cm are mounted 2 mm
apart in vacuum. The lower plate is earthen and the upper one is given a
positive charge of 0.05 . Find the electric field intensity and the potential

difference between the plates.

3
Ans: 112.9 × 10 V/m, 226 V



























































32 Electrostatics