Electric Charge Charge is the property associated with matter due to which it produces and experiences electric and magnetic effect.
 Conductors and Insulators Those substances which readily allow the passage of electricity through them are called conductors, e.g. metals, the earth and those substances which offer high resistance to the passage of electricity are called insulators, e.g. plastic rod and nylon.

The transference of electrons is the cause of frictional electricity.
 Additivity of Charges Charges are scalars and they add up like real numbers. It means if a system consists of n charges q1, q2, q3 , … ,qn, then total charge of the system will be q1 +q2 + … +qn.
 Conservation of Charge The total charge of an isolated system is always conserved, i.e. initial and final charge of the system will be same.

Quantization of Charge Charge exists in discrete amount rather than continuous value and hence, quantized.
Mathematically, charge on an object, q=±ne
where n is an integer and e is an electronic charge. When any physical quantity exists in discrete packets rather than in continuous amounts, the quantity is said to be quantized. Hence, the charge is quantized.

 Units of Charge
(i) SI unit coulomb (C)
(ii) CGS system
(a) electrostatic unit, esu of charge or stat-coulomb (stat-C)
(b) electromagnetic unit, emu of charge or ab-C (ab-coulomb)
1 ab-C = 10 C, 1 C = 3 x 109 stat-C

(a) Consider that the Universe is spherical with radius R made up of H atoms.

Charge on proton = \(-(1+y)e\)

So the total charge on H atom i.e.e \(_H={e_p+e}\)

\(=-(1+y)e+e\)

\(=[-1-y+1]e\)

Charge on 1 H atom \(=-ye\)

The number of H atoms in spherical Universe=N.V.

\(=N.\frac{4}{3} \pi R^3\)

\(\therefore\)The net charge in Universe \(=N.\frac{4}{3}\pi R^3 (-ye)\)

Consider the boundaries of Universe as Gaussian surface then by Gauss’s law of electrostatics,

\(\oint_s E.ds= \frac{q}{\varepsilon_0}\)

\(E.4 \pi R^2 = \frac{-4\pi NR^3ye}{3\in _0}\)

\(E=\frac{-4\pi NR^3 ye}{3\varepsilon _0.4\pi R^2}= \frac{-NRye}{3\varepsilon_0}\)

Electrostatic force or Columb force acting on one H atom \(F_C=qE\)

\(F_C= \frac{(-ye)(-NRye)}{3\varepsilon_0}=\frac{+y^2e^2NR}{3\varepsilon_0}\)

Positive sign \(F_C\) shows repulsive force.

We know, gravitational potential at boundary of Universe \(=\frac{GM}{R^2}\)

M=mass of Universe (or all H atoms)

So, the gravitational force acting on H atom at boundary of Universe= Gravitational potential \(\times\) m [m=mass of H-atom]

\(F_G=\frac{GM}{R^2}m_H\)

Mass of 1H atom = mass of a proton \(=m_p\)

\(\therefore\) Mass of Universe(M)= No. of H atoms in Universe \(\times m_p\)

\(=N.\frac{4}{3}\pi R^3 m_p\)

\(\therefore F_G =\frac{G\Big(N\frac{4}{3}\pi R^3\Big)m^2_p}{R^2}\)

\(\Rightarrow F_G= \frac{4\pi GNR \ m^2_p}{3}\)

If \(F_C\gt F_G\) then the universe will start to expand. So for critical value of expansion of y would be when

\(F_C=F_G\)

\(\frac{y^2e^2NR}{3\varepsilon_0}=\frac{4\pi G NR \ m^2_p}{3}\)

\(\Rightarrow y^2= \frac{\varepsilon \ \in_0 4\pi GNR m^2_p}{ \varepsilon \ e^2NR}=4\pi \in_0G \Big(\frac{m_p}{e}\Big)^2 \)

\(\Rightarrow y^2= \frac{6.67\times 10^{-11}}{9\times 10^9}\Big(\frac{1.66\times 10^{-27}}{1.6\times 10^{-19}}\Big)^2\)

\(\Rightarrow y^2 \cong \frac{6.67\times 10^{-20}}{9}[10^{-27+19}]^2\)

\(=0.741\times 10^{-20}\times (10^{-8})^2\)

\(\Rightarrow y^2=74.1\times 10^{-22}\times 10^{-16}\Rightarrow y= \sqrt{74.1\times 10^{-38}}\)

\(y=8.6\times 10^{-19} \simeq 10^{-18}\)

So critical value of y is of the order of \(10^{-18}\) so that Universe start to expand.

(b) For expansion repulsive force FC must be greater than, attractive gravitational force So net force on H atom to expand

\(F_H=F_C-F_G\)

\(\Rightarrow F_H=\frac{y^2e^2 NR}{3\varepsilon_0}-\frac{4\pi}{3}GNR m^2_p\)

This force \(F_H\) will produce acceleration in H atom

\(\therefore F_H=m_p \frac{d^2R}{dt^2}\)

Here R (size of Universe) changes with time as Universe expands with velocity

\(m_p\frac{d^2R}{dt^2}=\Bigg[\frac{Ny^2e^2}{3\varepsilon_0}-\frac{4\pi GNm^2_p}{3}\Bigg]R\)

\(\frac{d^2R}{dt^2}=\frac{1}{m_P} \Bigg[\frac{Ny^2e^2}{3\varepsilon_0}-\frac{4\pi GNm^2_p}{3}\Bigg]R\)

As \(N,y,e, \in_0,\pi, G,m_p\) are constants so taking a new constant \(\alpha^2\) such that

\(\alpha^2= \frac{1}{m_p}\Bigg[\frac{Ny^2e^2}{3 \in _0}-\frac{4\pi GNm^2_p}{3}\Bigg]\)

\(\frac{d^2 R}{dt^2}=\alpha^2 R\)

It is a differential equation of order 2. Its solution is

\(R=Ae^{\alpha t}+Be^{-\alpha t}\)

For expansion of Universe B=0

\(\therefore R= Ae^{\alpha t}\)

Receding velocity of Universe \(\nu= \frac{dR}{dt}\)

\(\nu=A\alpha e^{\alpha t}\) or \(\nu=R\alpha\)

As \(\alpha\) is constant. So the receding (expanding) velocity of Universe is directly proportional to the distance of matter (H atom) from the centre of Universe.

Electrostatic forces (Colombian forces) are conservative forces.
Principle of Superposition of Electrostatic Forces This principle states that the net electric force experienced by a given charge particle q0 due to a system of charged particles is equal to the vector sum of the forces exerted on it due to all the other charged particles of the system.

Electrostatic Force due to Continuous Charge Distribution
The region in which charges are closely spaced is said to have a continuous distribution of charge. It is of three types given as below:

Electric Field Intensity The electric field intensity at any point due to source charge is defined as the force experienced per unit positive test charge placed at that point without disturbing the source charge. It is expressed as

The charge on Cs atom \(=+e\)

The charge on Cl atom \(=- e\)

The distance of \(Cl^-\) ion from any \(Cs^+\) ion

\(=\frac{1}{2}\) diagonal of the cube of side 1

\(r= \frac{1}{2}\sqrt{3l^2}\)

\(=\frac{1}{2}\sqrt{3\times 0.4\times 10^{-9}\times 0.4\times 10^{-9}} m\)

\(=\frac{1}{2}\times 0.4\times 10^{-9}\times \sqrt{3}\ r=0.2\sqrt{3} \times 10^{-9} m\)

As the distance of \(Cl^-\) ion from \(Cs^+\) is equal and charge on each \(Cs^+\) ion is same, so electrostatic force due to \(Cs^+\) ion at A and D will be equal in magnitude and opposite in direction. So by \(F=\frac{kq_1q_2}{r^2}\) the net force on \(Cl^-\) due to \(Cs^+\) ion will be zero, as atoms of Cs attracts the \(Cl^-\) equally in opposite direction with pairs diagonally i.e., (B, H), (C, E), (D, F).

(ii) As the \(Cs^+\) ion at A is missing so the net force on \(Cl^-\) ion will be only due to its’ opposite \(Cs^+\) ion and other forces will be cancelled out.

Net forces on \(Cl^-\) ion when \(Cs^+\) ion from A is removed,

\(=\frac{kq_1q_2}{r^2}\Big(k=\frac{1}{4\pi \varepsilon_0}\Big)\)

\(r=0.2\sqrt{3} \times 10^{-9}\ m\) and \(|q_1|=|q_2|=|e|\)

\(e=1.6\times 10^{-19} C\)

\(F= \frac{9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.2\sqrt{3}\times 10^{-9}\times 0.2\sqrt{3}\times 10^{-9}}\)

\(=\frac{9\times 16\times 16 \times10^{-38+9}}{2\times 2\times 3\times 10^{-18}}\)

\(=3\times16\times 4\times 10^{-29+18}=192\times 10^{-11}\)

Force on \(Cl^-\) ion

\(F=1.92\times 10^{-9} N\) towards \(Cl^-\) ion

Electric Field due to a System of Charges
Same as the case of electrostatic force, here we will apply the principle of superposition, i.e.

 Electric Field Lines Electric field lines are a way of pictorially mapping the electric field around a configuration of charge(s). These lines start on a positive charge and end on a negative charge. The tangent on these lines at any point gives the direction of the field at that point

Electric field lines due to positive and negative charges and their combinations are shown as below:

 Electric Dipole Two point charges of the same magnitude and opposite nature separated by a small distance altogether form an electric dipole.
 Electric Dipole Moment The strength of an electric dipole is measured by a vector quantity known as electric dipole moment (p) which is the product of the charge (q) and separation between the charges (2l).
 

 Electric Field due to a Dipole Electric field of an electric dipole is the space around the dipole in which the electric effect of the dipole can be experienced.

 Dipole is in stable equilibrium in uniform electric field when angle between p and E is 0° and in unstable equilibrium when angle θ= 180°.
Net force on electric dipole placed in a uniform electric field is zero.
 There exists a net force and torque on electric dipole when placed in non-uniform electric field.
 Work done in rotating the electric dipole from θ1 to θ2 is W = pE (cos θ1 – cos θ2)
 Potential energy of electric dipole when it rotates from θ1 = 90° to θ2 =0
U = pE (cos 90° – cosθ) = -pE cos θ = – p .E

 Work done in rotating the dipole from the position of stable equilibrium to unstable equilibrium, i.e. when θ1 = 0° and θ2 = π.
W = 2 pE
Work done in rotating the dipole from the position of stable equilibrium to the position in which dipole experiences maximum torque, i.e. when θ1 = 0° and θ2 = 90°.
W = pE