Chapter Notes: Electrostatics - Class 12 Physics Notes

Notes for Electrostatics chapter of class 12 physics. Dronstudy provides free comprehensive chapterwise class 12 physics notes with proper images & diagram.


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Charge is the property of matter that causes it to produce and experience electrical and magnetic effects. The study of the electrical charges at rest is called electrostatics. When both electrical and magnetic effects are present, the interaction between charges is referred to as electromagnetic.

There exist two types of charges in nature : positive and negative. Like charges repel, and unlike charges attract, each other.

The type of charge on an electron is negative. The charge of a proton is the same as that of an electron but with a positive sign. In an atom, the number of electrons and the number of protons are equal. The atom is, therefore, electrically neutral. If one or more electrons are added to it, it becomes negatively charged and is designated as negative ion. However, if one or more electrons are removed from an atom, it becomes positively charged and is called a positive ion.

The excess or deficiency of electrons in a body gives the concept of charge. If there is an excess of electrons in a body, it is negatively charged. And if there is deficiency of electrons, the body becomes positively charged. Whenever addition or removal of electrons takes places, the body acquires a charge.

The SI Unit of charge is coulomb (C). In SI units, the current is a fundamental quantity, having a unit of ampere (A). The unit of charge is defined in terms of the unit of current. Thus, one coulomb is the charge transferred in one second across the section of a wire carrying a current of one ampere.

      As    q = It, we have
1 C = (1 A) (1 s)

      The dimensions of charge are [A T].

Properties of Charge

(1)  Quantization of Charge :  Electric charge can have only discrete values, rather than any value. That is, charge is quantized. The smallest discrete value of charge that can exist in nature is the charge on an electron, given as

e =  ± 1.6 x 10- 19 C

This is the charge attained by an electron and a proton.
A charge q must be an integral multiple of this basic unit. That is,

Q = ± ne          where n = 1, 2, …

Charge on a body can never be (½)e, (2/3)e, or 5.7e, etc.
When we rub a glass rod with silk, some electrons are transferred from the rod to the silk. The rod becomes positively charged. The silk becomes negatively charged. The coulomb is a very large amount of charge. A typical charge acquired by a rubbed body is 10 - 8 C.

Application  1
A body is having a charge of +0.32 C. How many electrons have been added to or removed from it ?


Given q = +0.32 C. Since the charge is positive, there is deficiency of electrons.

n = {q \over e} = {{0.32} \over {1.6 \times {{10}^{ - 19}}}} = 2 x 1018 electrons

Note that the electron itself is not the charge; charge is a property, like mass, of elementary particles, such as the electrons, protons, etc.

(2) Charge is Always Associated with Mass : A charge cannot exist without mass, though a mass can exist without charge. The particles such as photon or neutrino have no (rest) mass. Hence, these particles can never have a charge.
The mass of a body (slightly) increases when it acquires a negative charge (by gaining some electrons). On the other hand, when a body acquires a positive charge (by losing some electrons), its mass (slightly) decreases.

(3) Conservation of Charge : In an isolated system, the total charge remains constant. In other words, charge can neither be created nor destroyed. It can be transferred from one body to the other. Or, equal amounts of positive or negative charges can appear or disappear. This is what happens in pair production and pair annihilation, as shown in figure.

Note that in pair production and pair annihilation, neither mass nor energy is conserved separately, but (mass + energy) is conserved. In pair production energy is converted into mass, while in annihilation mass is converted into energy.
Conservation of charge holds good in all types of reactions.
For example :
Chemical Reaction :

                Na  +     Cl-   \to    NaCl
   Charge :   (+e)   +   (-e)   =  (0)

Radioactive Decay :

                                            n          \to      p          +          e-             +           {\bar v}

                                    Neutron                proton              electron         antineutrino

Charge :          (0)        =       (+e)      +       (-e)          +         (0)

(4) Invariance of Charge :  Numerical value of a charge is independent of the frame of reference. It means the value of charge on a body remains the same, whether it is stationary, or moving with a constant velocity or accelerating. In contrast, the mass of a body depends on its speed, and it decreases with increase in speed.


The force of interaction of two stationary point charges in vacuum is directly proportional to the product of these charges and inversely proportional to the square of their separation,

F = {{k{q_1}{q_2}} \over {{r^2}}}

where F is in newton, q1 and q2 in coulomb, r in metre, and k is a constant given in SI units by

k{\rm{ }} = {\rm{ }}{{\rm{1}} \over {{\rm{4}}\pi {\rm{ }}{ \in _{\rm{0}}}}} = 9 x 109 N m2 C-2

where { \in _0} = 8.85 ´ 10-12 C2 N-1 m-2 and is called the permittivity of free space (vacuum or air).
For mediums other than air or vacuum, the electrostatic force between two charges becomes

F = {1 \over {4\pi \in }}.{{{q_1} {q_2}} \over {{r^2}}} = {1 \over {4\pi{\varepsilon _0}{ \in _r}}}.{{{q_1} {q_2}} \over {{r^2}}}

Here  \in = { \in _0}{ \in _r}, is called the absolute permittivity or permittivity of the medium, and { \in _r} = { \in \over {{ \in _0}}} is the relative permittivity of the medium which is a dimensionless constant. { \in _r} is also sometimes called dielectric constant, and is represented by letter K.

The coulomb force acts along the straight line connecting the points of location of the charges.
This force is central and spherically symmetric.
The vector form of Coulomb’s law is {\rm{\vec F }} = {\rm{ }}{{k{q_1}{q_2}} \over {{r^2}}}{\rm{ \hat r}}. The unit vector  has its origin at the source of the force.
For example, to find the force on q2, the origin of r is placed at q1 as shown in the figure. If F is the magnitude of the force (a positive scalar), then
\vec F = + F\hat r          means a repulsion
whereas,  \vec F = - F\hat r            means an attraction

Analogy with Gravitation Law

Coulomb’s law is analogous to Newton’s law of gravitation :

F = G {{{m_1}{m_2}} \over {{r^2}}}

However, following are the important differences :
(a)  Electric force between charged particles is much stronger than gravitational force, i.e., FE >> FG. This is why when both FE and FG are present, we neglect FG
(e.g. between two electrons FG = 10-39 FE).
(b)  Electric force can be attractive or repulsive. But the gravitational force is always attractive.
(c)  Electric force depends on the nature of medium between the charges, but gravitational force does not.

Important Points Regarding Coulomb’s Law

(1)  Charges are Assumed to be at Rest :  When charges are in motion they also produce and experience magnetic forces.

(2) Charges are Assumed to be on Point Particles : Coulomb’s law cannot be directly applied to a finite charge distribution. In such a case (see figure), it is not possible to definitely specify the separation between the charges. However, there is an exception. When the charge is distributed uniformly over a spherical surface, the force on a point charge outside the surface may be computed from Coulomb’s law by treating the charge on the sphere as if it were concentrated at the centre.

Principle of Superposition

The coulomb’s law obeys the principle of superposition. It means that the force between two particles is not affected by the presence of other charges. This principle is used to find the net force exerted on a given charged particle by other charged particles.
The force on a charged particle q1 due to point charges q2, q3 and q4 is the resultant of forces due to individual point charges, i.e.,
{\vec F_1} = {\vec F_{12}} + {\vec F_{13}} + {\vec F_{14}}
Note that the notation represents the force on q1 due to q­2.

How to Solve a Problem using Coulomb’s Law ?

(1)  Decide whether the force due to a given charge is attractive or repulsive and show it by drawing vector, pointing towards or away from the given charge, respectively.
(2)  Find the magnitude of the force using Coulomb’s law---ignoring the signs of the charges.
(3)  Resolve the forces along the given co-ordinate axes and express them in vector form using \hat i,\hat j,\hat k unit vector notation, unless otherwise specified.
(4)  Use the principle of superposition to find the net force on the charge.

Application 2
Four point charges are located at the corners of a rectangle, as shown in figure. Find the net force acting on the charge q1

Solution:(1)  The force {\vec F_{12}} (between q1 and q2) is repulsive, while the forces {\vec F_{13}} (between q1 and q3) and {\vec F_{14}} (between q1 and q4) are attractive.

(2) The magnitude of the forces {\vec F_{12}}, {\vec F_{13}} and {\vec F_{14}} are

{F_{12}} = {{k{q_1}{q_2}} \over {r_{12}^2}} = {{\left( {9 \times {{10}^9}} \right) \left( {8 \times {{10}^{ - 9}}} \right) \left( {4 \times {{10}^{ - 9}}} \right)} \over { {{\left( {3 \times {{10}^{ - 2}}} \right)}^2}}} = 32 \times {10^{ - 5}}N
{F_{13}} = {{k{q_1}{q_3}} \over {r_{13}^2}} = {{\left( {9 \times {{10}^9}} \right) \left( {8 \times {{10}^{ - 9}}} \right) \left( {8 \times {{10}^{ - 9}}} \right)} \over { {{\left( {5 \times {{10}^{ - 2}}} \right)}^2}}} = 23 \times {10^{ - 5}}N
{F_{14}} = {{k{q_1}{q_4}} \over {r_{14}^2}} = {{\left( {9 \times {{10}^9}} \right) \left( {8 \times {{10}^{ - 9}}} \right) \left( {12 \times {{10}^{ - 9}}} \right)} \over { {{\left( {4 \times {{10}^{ - 2}}} \right)}^2}}} = 54 \times {10^{ - 5}}N

(3)  {\vec F_{12}}  = (- 32 \times {10^{ - 5}})\hat j

{\vec F_{13}}  = ({F_{13}}cos\theta ) \hat i +  + ({F_{13}}sin\theta ) \hat j
 = \left[ { - {\rm{(23) }}\left( {{{\rm{4}} \over {\rm{5}}}} \right){\rm{ }}\hat i{\rm{ }} + {\rm{ (23) }}\left( {{{\rm{3}} \over {\rm{5}}}} \right){\rm{ }}\hat j} \right] \times {\rm{ 1}}{{\rm{0}}^{{\rm{ - 5}}}}
 = [ - 18.4\hat i + 13.8\hat j] \times {10^{ - 5}}
{\vec F_{14}}  = ( - 54 \times {10^{ - 5}}) {\hat i}

(4)  The net force on q1 is

{\vec F_1} = {\vec F_{12}} + {\vec F_{13}} + {\vec F_{14}}
 = [ - 72.4\hat i - 18.2\hat j] \times {10^{ - 5}}N
 = [ - 72.4 - 18.2] \times {10^{ - 5}}N

Application 3
Five point charges, each +q are placed on five vertices of a regular hexagon of side L. What is the magnitude of the force on a point charge –q placed at the centre of the hexagon ?


Had there been sixth charge +q at the remaining vertex of hexagon, the net force due to all the six charges on –q at O would be zero. The forces due to individual charges  will balance each other. That is,
{\overrightarrow {\rm{F}} _{\rm{R}}} = 0
Now if {\rm{\vec f}} is the force due to sixth charge and due to remaining five charges, we must have
{{\rm{\vec F}}_R} = {\rm{\vec F}} + {\rm{\vec f}} = 0  i.e.,          {\rm{\vec F }} = - {\rm{\vec f}}
or      F = {1 \over {4\pi {\varepsilon _0}}} {{q \times q} \over {{L^2}}} = {\rm{ }}{1 \over {4\pi {\varepsilon _0}}} {\left[ {{q \over L}} \right]^2}


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