(a) This is because the comb gets charged by friction. The molecules in the paper gets polarised by the charged comb, resulting in a net force of attraction. If the hair is wet, or if it is rainy day, friction between hair and the comb reduces. The comb does not get charged and thus it will not attract small bits of paper.
(b) To enable them to conduct charge (produced by friction) to the ground; as too much of static electricity accumulated may result in spark and result in fire.
(c) Reason similar to (b).
(d) Current passes only when there is difference in potential

(a) \(U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r}=9 \times 10^{9} \times \frac{7 \times(-2) \times 10^{-12}}{0.18}=-0.7 \mathrm{~J}\)
(b) W = U₂ – U₁ = 0 – U = 0 – (–0.7) = 0.7 J.
(c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find
\(q_{1} V\left(\mathbf{r}_{1}\right)+q_{2} V\left(\mathbf{r}_{2}\right)=A \frac{7 \mu \mathrm{C}}{0.09 \mathrm{~m}}+A \frac{-2 \mu \mathrm{C}}{0.09 \mathrm{~m}}\)
and the net electrostatic energy is
\(q_{1} V\left(\mathbf{r}_{1}\right)+q_{2} V\left(\mathbf{r}_{2}\right)+\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r_{12}}=A \frac{7 \mu C}{0.09 m}+A \frac{-2 \mu C}{0.09 m}-0.7 \mathrm{~J}\)
= 70 − 20 − 0.7 = 49.3 J

(a) As \(V \propto \frac{1}{r}, V_{P}>V_{Q^{}}\) Thus, (V_P – V_Q) is positive. Also V_B is less negative than V_A . Thus, V_B > V_A or (V_B – V_A) is positive.
(b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly, (P.E.)_A > (P.E.)_B and hence sign of potential energy differences is positive.
(c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative.
(d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive.
(e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A.

ABCDEF is a regular hexagon of side 10 cm each. At each corner, the charge q =5 \(\mu\)C is placed. O is the centre of the hexagon.

Given, AB = BC = CD = DE
= EF = FA = d = 10 cm
As, the hexagon has six equilateral triangles, so the distance of centre O from every vertex is 10 cm.
i.e. OA = OB = OC = OD
= OE = OF = d = 10 cm
\(\therefore\) Potential at point O = Sum of potentials at centre O due to individual point charge
i.e. V_O = V_A + V_B + V_C + V_D + V_E + V_F
\(=\frac{1}{4\pi \varepsilon_{0}}.\left [ \frac{q}{OA}+\frac{q}{OB}+\frac{q}{OC}+\frac{q}{OD}+\frac{q}{OE}+\frac{q}{OF} \right ]\)
\(=\frac{1}{4\pi \varepsilon _{0}}.\frac{6q}{d}\)    \(\left [ \because V=\frac{1}{4\pi \varepsilon _{0}.\frac{q}{r}} \right ]\)
Putting the values, we get
\(=9\times 10^{9}\times \frac{6\times 5 \times 10^{-6}}{10\times 10^{-2}}\)
= 2.7 \(\times\)10⁶ V

Given,
q₁ = 5 x 10^-8 C
q₂ = -3 x 10^-8 
The two charges are at a distance, d = 16cm = 0.16m from each other.
Let us consider a point “P” over the line joining charges q₁ and q₂.
Let the distance of the considered point P from q₁ be ‘r
Let us consider point P to have zero electric potential (V)
The electric potential at point P is the summation of potentials due to charges q₁ and q₂.
Therefore \(V=\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q_{1}}{r}+\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q_{2}}{d-r}\)     ..........(i)
Here,
ϵ_o​ = permittivity of free space.
Putting  V = 0, in eqn. (1), we get,
\(0=\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q_{1}}{r}+\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q_{2}}{d-r} \quad \frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q_{1}}{r}=-\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q_{2}}{d-r} \quad \frac{q_{1}}{r}=-\frac{q_{2}}{d-r} \quad \frac{5 \times 10^{-8}}{r}=-\frac{\left(-3 \times 10^{-8}\right)}{0.16-r}\)
5(0.16 – r)  = 3r
0.8 = 8r
r = 0.1m = 10 cm.
Therefore, at a distance of 10 cm from the positive charge, the potential is zero between the two charges.
Let us assume a point P at a distance ‘s’  from the negative charge be outside the system, having potential zero.
So, for the above condition, the potential is given by –
\(V=\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q_{1}}{s}+\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q_{2}}{s-d}\)       ...........(ii)
Here,
ϵ_o​ = permittivity of free space.
For V = 0, eqn. (2) can be written as :
\(0=\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q_{1}}{s}+\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q_{2}}{s-d} \quad \frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q_{1}}{s}=-\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q_{2}}{s-d} \quad \frac{q_{1}}{s}=-\frac{q_{2}}{s-d} \quad \frac{5 \times 10^{-8}}{s}=-\frac{\left(-3 \times 10^{-8}\right)}{s-0.16}\)
5(s – 0.16) = 3s
0.8  = 2s
S = 0.4 m = 40 cm.
Therefore, at a distance of 40 cm from the positive charge outside the system of charges, the potential is zero.

(a) The charge on the capacitor is
Q = CV = 900 x 10^–12 F x 100 V = 9 x 10^–8 C
The energy stored by the capacitor is
= (1/2) CV² = (1/2) QV
= (1/2) x 9 x 10^–8C x 100 V = 4.5 x 10^–6 J
(b) In the steady situation, the two capacitors have their positive plates at the same potential, and their negative plates at the same potential. Let the common potential difference be V′. The charge on each capacitor is then Q′ = CV′. By charge conservation, Q′ = Q/2. This implies V′ = V/2. The total energy of the system
\(=2 \times \frac{1}{2} Q^{\prime} V^{\prime}=\frac{1}{4} Q V=2.25 \times 10^{-6} \mathrm{~J}\)
Thus in going from (a) to (b), though no charge is lost; the final energy is only half the initial energy.
There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in the form of heat and electromagnetic radiation

(a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A, and then the charges –q, +q, and –q are brought to B, C and D, respectively. The total work needed can be calculated in steps:
(i) Work needed to bring charge +q to A when no charge is present elsewhere: this is zero.
(ii) Work needed to bring –q to B when +q is at A. This is given by (charge at B) x (electrostatic potential at B due to charge +q at A)
\(=-q \times\left(\frac{q}{4 \pi \varepsilon_{0} d}\right)=-\frac{q^{2}}{4 \pi \varepsilon_{0} d}\)
(iii) Work needed to bring charge +q to C when +q is at A and –q is at B. This is given by (charge at C) x (potential at C due to charges at A and B)
\(=+q\left(\frac{+q}{4 \pi \varepsilon_{0} d \sqrt{2}}+\frac{-q}{4 \pi \varepsilon_{0} d}\right)\)
\(=\frac{-q^{2}}{4 \pi \varepsilon_{0} d}\left(1-\frac{1}{\sqrt{2}}\right)\)
(iv) Work needed to bring –q to D when +q at A,–q at B, and +q at C. This is given by (charge at D) x (potential at D due to charges at A, B and C)
\(=-q\left(\frac{+q}{4 \pi \varepsilon_{0} d}+\frac{-q}{4 \pi \varepsilon_{0} d \sqrt{2}}+\frac{q}{4 \pi \varepsilon_{0} d}\right)\)
\(=\frac{-q^{2}}{4 \pi \varepsilon_{0} d}\left(2-\frac{1}{\sqrt{2}}\right)\)
Add the work done in steps (i), (ii), (iii) and (iv). The total work required is
\(=\frac{-q^{2}}{4 \pi \varepsilon_{0} d}\left\{(0)+(1)+\left(1-\frac{1}{\sqrt{2}}\right)+\left(2-\frac{1}{\sqrt{2}}\right)\right\}\)

\(=\frac{-q^{2}}{4 \pi \varepsilon_{0} d}(4-\sqrt{2})\)
The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.
(Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.)
(b) The extra work necessary to bring a charge q₀ to the point E when the four charges are at A, B, C and D is q₀ x (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by thatdue to B and D. Hence no work is required to bring any charge to  point E.

Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis; the negative charge is taken to be on the right side of the origin

Let P be the required point on the x-axis where the potential is zero. If x is the x-coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for x < 0.) If x lies between O and A, we have
\(\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{3 \times 10^{-8}}{x \times 10^{-2}}-\frac{2 \times 10^{-8}}{(15-x) \times 10^{-2}}\right]=0\)
where x is in cm. That is,
\(\frac{3}{x}-\frac{2}{15-x}=0\)
which gives x = 9 cm
If x lies on the extended line OA, the required condition is
\(\frac{3}{x}-\frac{2}{x-15}=0\)
which gives
x = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.

(a) In the given network, C_1, C₂ and C₃ are connected in series. The effective capacitance C′ of these three capacitors is given by
\(\frac{1}{C^{\prime}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}\)
For C₁ = C₂ = C₃ = 10 μF, C′ = (10/3) μF. The network has C′ and C₄ connected in parallel. Thus, the equivalent capacitance C of the network is
\(C=C^{\prime}+C_{4}=\left(\frac{10}{3}+10\right) \mu F=13.3 \mu \mathrm{F}\)
(b) Clearly, from the figure, the charge on each of the capacitors, C₁, C₂ and C₃ is the same, say Q. Let the charge on C₄ be Q′. Now, since the potential difference across AB is Q/C₁, across BC is Q/C₂, across CD is Q/C₃ , we have
\(\frac{Q}{C_{1}}+\frac{Q}{C_{2}}+\frac{Q}{C_{3}}=500 \mathrm{~V}\)
Also, Q′/C₄ = 500 V.
This gives for the given value of the capacitances
\(Q=500 \mathrm{~V} \times \frac{10}{3} \mu \mathrm{F}=1.7 \times 10^{-3} \mathrm{C}\)  and
Q′ = 500 V x 10 μF = 5.0 x10⁻³ C

\(V=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2} \times \frac{4 \times 10^{-7} \mathrm{C}}{0.09 \mathrm{~m}}\)
= 4 x 10⁴ V 
(b) W = qV = 2 x 10⁻⁹C x 4 x 10⁴ V
= 8 x 10^–5 J
No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along r and another perpendicular to r. The work done corresponding to the later will be zero.

Given, C₁ = C₂ = 600 pF
= 600 \(\times\)10^-12F
= 6 \(\times\)10^-10 F
V₁ = 200 V, V₂ = 0
\(\begin{aligned} \therefore \text { Energy lost } & =\frac{C_1 C_2\left(V_1-V_2\right)^2}{2\left(C_1+C_2\right)} \\ \end{aligned}\)
\(\begin{aligned} =\frac{\left(6 \times 10^{-10}\right)^2(200-0)^2}{2 \times 12 \times 10^{-10}} \end{aligned}\)
= 6 \(\times\)10^-6 J

(a) Dielectric constant of the mica sheet, k = 6
If voltage supply remained connected, the voltage between two plates will be constant.
Supply voltage, V = 100 V
Initial capacitance, C = 1.771 x 10⁻¹¹ F
New capacitance, C₁ = kC = 6 x 1.771 x 10⁻¹¹ F = 106 pF
New charge, q₁ = C₁V = 106 x 100 pC = 1.06 x 10^–8 C
Potential across the plates remain 100 V.
(b) Dielectric constant, k = 6
Initial capacitance, C = 1.771 x 10⁻¹¹ F
New capacitance, C₁ = kC = 6 x 1.771 x 10⁻¹¹ F = 106 pF
If the supply voltage is removed, then there will be a constant amount of charge in the plates.
Charge = 1.771 x 10⁻⁹ C
Potential across the plates is given by,
\(\mathrm{V} 1=\mathrm{q} / \mathrm{C} 1=\frac{1.771 \times 10^{-9}}{106 \times 10^{-12}}\)
= 16.7 V

(1) Given, C₁ = 2pF, C₂ = 3pF and C₃ = 4pF.
Equivalent capacitance for the parallel combination is given by C_eq .
Therefore, C_eq = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9pF
Hence, the total capacitance of the combination is 9pF.
(2) Supply voltage, V = 100V
The three capacitors are having the same voltage, V = 100v
q = VC
where,
q = charge
C = capacitance of the capacitor
V = potential difference
for capacitance, c = 2pF
q = 100 x 2 = 200pC = 2 x 10^-10C
for capacitance, c = 3pF
q = 100 x 3 = 300pC = 3 x 10^-10C
for capacitance, c = 4pF
q = 100 x 4 = 400pC = 4 x 10^-10 C

Given,
Capacitance, C = 8pF.
In the first case, the parallel plates are at a distance ‘d’  and is filled with air.
Air has dielectric constant, k = 1
Capacitance, C\(=\frac{k \times \epsilon_{o} \times A}{d}=\frac{\epsilon_{o} \times A}{d}\)          .......(i)
Here,
A = area of each plate
ϵ_o​ = permittivity of free space.
Now, if the distance between the parallel plates is reduced to half, then d₁ = d/2
Given, dielectric constant of the substance, k₁ = 6
Hence, the capacitance of the capacitor,
\(\mathrm{C}_{1}=\frac{k_{1} \times \epsilon_{o} \times A}{d_{1}}=\frac{6 \epsilon_{0} \times A}{d / 2}=\frac{12 \epsilon_{o} A}{d}\)         .......(ii)
Taking ratios of eqns. (1) and (2), we get,
C₁ = 2 x 6 C = 12 C = 12 x  8 pF = 96pF.
Hence, the capacitance between the plates is 96pF.

(1) Given,
Radius of spherical conductor, r = 12cm = 0.12m
Charge is distributed uniformly over the surface, q = 1.6 x 10^-7 C.
The electric field inside a spherical conductor is zero.
(2) Electric field E, just outside the conductor is given by the relation
\(\mathrm{E}=\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q}{r^{2}}\)
Here, permittivity of free space and \(\frac{1}{4 \pi \epsilon_{o}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\)
Therefore,
\(\mathrm{E}=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{(0.12)^{2}}=10^{5} \mathrm{NC}^{-1}\)
Therefore, just outside the sphere the electric field is 4.4 x 10⁴ NC^-1.
(3) From the centre of the sphere the electric field at a point 18m = E₁.
From the centre of the sphere, the distance of point d = 18 cm = 0.18m
\(\mathrm{E}_{1}=\frac{1}{4 \pi \epsilon_{o}} \cdot \frac{q}{d^{2}}=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{\left(1.8 \times 10^{-2}\right)^{2}}=4.4 \times 10^{4} \mathrm{NC}^{-1}\)
So, from the centre of sphere the electric field at a point 18 cm away is 4.4 x 10⁴ NC^-1.

(1) An equipotential surface is defined as the surface over which the total potential is zero. In the given question the plane is normal to line AB. The plane is located at the mid – point of the line AB as the magnitude of the charges are same.
(2) At every point on this surface the direction of the electric field is normal to the plane in the direction of AB.

Given,
Capacitance of the capacitor, C = 12pF = 12 x 10^-12 F
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
\(\mathrm{E}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 12 \times 10^{-12} \mathrm{\times}(50)^{2} \mathrm{~J}=1.5 \times 10^{-8} \mathrm{~J}\)
Therefore, the electrostatic energy stored in the capacitor is 1.5 x 10^-8 J. was disconnected.

Given,
The area of plate of the capacitor, A = 6 x 10^-3 m²
Distances between the plates, d = 3mm = 3 x 10^-3 m
Voltage supplied, V = 100V
Capacitance of a parallel plate capacitor is given by, \(C=\frac{\epsilon \times A}{d}\) 
Here,
ε = permittivity of free space = 8.854 x10^-12 N^-1 m ^-2 C^-2
\(C=\frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}=17.81 \times 10^{-12} \mathrm{~F}=17.71 \mathrm{pF}\)
Therefore, each plate of the capacitor is having a charge of
q = VC = 100 x 17.81 x 10^-12 C = 1.771 x 10^-9 C

There are three capacitors cach of capacitance 9 pF.
\(\therefore\) C₁ = C₂ = C₃ = 9 pF
and voltage, V = 120 V
(i) The total capcitance in series combination,
\(\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\)
\(\Rightarrow \frac{1}{C_{s}}=\frac{3}{9} \Rightarrow C_{s}=3 pF\)
(ii) Let the charge across the system be q and potentials across C₁, C₂ and C₃ be V₁, V₂ and V₃, respectively.
Charge, q = C_s. V = 3 \(\times\)120 = 360 pC
Potential difference across C₁,
\(V_{1}=\frac{q}{C_{1}}=\frac{360}{9}=40 V\)
Potential difference across C₂,
\(V_{2}=\frac{q}{C_{2}}=\frac{360}{9}=40 V\)
Potential difference across C₃,
\(V_{3}=\frac{q}{C_{3}}=\frac{360}{9}=40 V\)
Thus, the potential difference across each capacitor is 40 V.