(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

(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.

(i) (b) Given, focal length of thin converging lens, f_c = 20 cm = f₁
Focal length of thin diverging lens, f_d = -15 cm = f₂
Power of the combination,  P = ?
By using the formula, \(\begin{aligned} \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2} \end{aligned}\)
\(\begin{aligned} \frac{1}{f}=\frac{1}{20}-\frac{1}{15}=\frac{3-4}{60}=\frac{-1}{60} \end{aligned}\)
\(\therefore\) f = -60
Power of lenses, \(P=\frac{100}{f}=\frac{100}{-60}\)
\(P=-\frac{5}{3} \mathrm{D}\)
(ii) (c) Given, radii of curvature of two surfaces of convex lens are R and 2R.
Focal length of the lens,
\(\begin{aligned} f=\frac{4}{3} R \end{aligned}\)
\(\begin{aligned} \mu=? \end{aligned}\)
We know that,
\(\begin{aligned} \frac{1}{f} & =(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \end{aligned}\)
\(\begin{aligned} \frac{3}{4 R} & =(\mu-1)\left(\frac{1}{R}+\frac{1}{2 R}\right) \end{aligned}\)
\(\begin{aligned} 3 & =(\mu-1) 4 R\left(\frac{3}{2 R}\right) \end{aligned}\)
\(\begin{aligned} 3 & =6 \mu-6 \end{aligned}\)
\(\begin{aligned} -6 \mu & =-6-3=-9 \end{aligned}\)
Refractive index, \(\mu=\frac{3}{2}\)
(iii) (a) The focal length of an equiconvex lens increases when the lens is dipped in water because the value of refractive index decreases.
(iv) (a) 

For convex lens,
\(\begin{gathered} u=\infty, f=10 \mathrm{~cm} \end{gathered}\)
\(\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \end{gathered}\)
\(\begin{gathered} \frac{1}{10}=\frac{1}{v}-\frac{1}{\infty} \end{gathered}\)
u = 10 cm
The image l₁, is formed in front of the mirror, so this image acts as object for a Concave mirror.
For concave mirror. u = -30 cm
f = -15 cm
\(\begin{aligned} \frac{1}{f} & =\frac{1}{v}-\frac{1}{u} \end{aligned}\)
\(\begin{aligned} \Rightarrow \frac{1}{-15} & =\frac{1}{v}-\frac{1}{-30} \end{aligned}\)
\(\begin{aligned} \frac{1}{v} & =\frac{-3}{30} \end{aligned}\)
\(\Rightarrow\)   v = -10 cm
The final image will be formed at a distance of 10 cm. left of the lens.
Or
(iv) (a) Focal length of convex lens L₁, f₁ = 16 cm
Focal length of convex lens L₂, f₂ = 12 cm
The coaxially distance, d = 40 cm
We know that, combined focal length,
\(\begin{aligned} \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \end{aligned}\)
\(\begin{aligned} \frac{1}{v} & =\frac{1}{f}+\frac{1}{u} \end{aligned}\)

\(\frac{1}{v}=\frac{1}{16}+\frac{1}{-\infty}\)
v = 16 cm
Positive sign shows that the image is formed to the right of the lens,
So this image acts as object for a convex lens I₂.
For this, f = 12 cm
\(\therefore\) u = (40 - 16) cm
 = 24 cm
u will taken with negative sign.
\(\begin{aligned} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \end{aligned}\)
\(\begin{aligned} \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \end{aligned}\)
\(\begin{aligned} \frac{1}{v}=+\frac{1}{12}-\frac{1}{24} \end{aligned}\)
\(\Rightarrow\)  v = + 24 cm (real)

(i) When air molecules/atoms are subjected to high electric field, the centres of negative charges (electrons) and positive charges (nucleus) are separated. Electrons experience force opposite to the direction of electric field and positive charge experiences force in the direction of electric field. If electric field is sufficiently high to pull out electrons, then air gets ionised and starts conducting.
(ii) \(\therefore \ \text { Electric field } \mathrm{E}=\frac{\mathrm{Q}}{4 \pi \varepsilon_{0}}\)
and E = 3 x 10 ⁶ V/m
then \(\mathrm{Q}=4 \pi \varepsilon_{0} \times 3 \times 10^{6} \mathrm{C}\)
\(\mathrm{Q}=\frac{3 \times 10^{6}}{9 \times 10^{9}}=0.33333 \times 10^{-3} \mathrm{C}\)
Q = 333.33 \(\mu\)C
(iii) If air is humid, then also it starts conducting and electric charge leaks away.

(i) variation of Electric field with distance
If x = 1 cm = 10^{- 2} m, then net Electric field at P
here r_A = 10^{- 2} m; r_B = 9 x 10^{- 2} m
\(\therefore \quad \mathrm{E}_{1}=9 \times 10^{9} \times 10^{-9}\left[\frac{1}{\left(10^{-2}\right)^{2}}-\frac{1}{\left(9 \times 10^{-2}\right)^{2}}\right]\)
q_A = q_B = 10^{- 9} C
k = 9 x 10^{- 9} Nm² C^-2
E₁ = 8.9 x 10 ⁴ N/C
Similarly if x = 2 cm
\(\mathrm{E}_{2}=9\left[\frac{1}{\left(2 \times 10^{-2}\right)^{2}}-\frac{1}{\left(8 \times 10^{-2}\right)^{2}}\right]\)
= 2.0 x 10 ⁴ N/C
for x = 3 cm
\(\mathrm{E}_{3}=9\left[\frac{1}{\left(3 \times 10^{-2}\right)^{2}}-\frac{1}{\left(7 \times 10^{-2}\right)^{2}}\right]\)
= 0.82 x 10⁴N/C
for x = 5 cm; E₅ = 0
for x = 7 cm
E₇ = - 0.82 x 10⁴ N/S
x = 8 cm, E₈ = -2 x 10^-4 N/C
x = 9 cm, E₉ = -8.9 x 10^-4 N/C

(ii) Variation of electric Potential is x
if x = 1 cm
\(\mathrm{V}_{1}=\frac{k q}{r_{\mathrm{A}}}+\frac{k q}{r_{\mathrm{B}}}\)
\(=k q\left[\frac{1}{r_{\mathrm{A}}}+\frac{1}{r_{\mathrm{B}}}\right]\)
\(\therefore \ \mathrm{V}_{1}=9 \times 10^{9} \times 10^{-9}\left[\frac{1}{10^{-2}}+\frac{1}{9 \times 10^{-2}}\right]\)
\(\mathrm{V}_{1}=9 \times 10^{2} \times \frac{10}{9}\)
= 1 x 10³ V
for x = 2 cm
\(\mathrm{V}_{2}=9\left[\frac{1}{2 \times 10^{-2}}+\frac{1}{8 \times 10^{-2}}\right]\)
\(=\frac{9 \times 10^{2} \times 10}{16}=0.56 \times 10^{3} \mathrm{~V}\)
x = 3 cm;
\(\mathrm{V}_{3}=9\left[\frac{1}{3 \times 10^{-2}}+\frac{1}{7 \times 10^{-2}}\right]\)
= 0.43 x 10³ ​​​​​​V
x = 5 cm;
\(\mathrm{V}_{5}=9 \times \frac{2 \times 10^{3}}{5 \times 10^{-2}}\)
= 0.36 x 10³ V
x = 7 cm;
Y₂ = 0.43 x 10³ V
x = 8 cm;
V₈ = 0.56 x 10³ V
X = 9 cm;
V₉ = 1.0 x 10³V

(iii) Electric field and potential at mid point (x = 5 cm) are 0 and 3.6 x 10³ V respectively .

(i) \(\mathrm{U}_{\mathrm{A}}=\frac{1}{2} \mathrm{C}_{\mathrm{A}} \mathrm{V}^{2}\)
\(=\frac{1}{2} \times\left(2 \times 10^{-6}\right) \times(100)^{2}\)
= 10^{- 2} J
\(\mathrm{U}_{\mathrm{B}}=\frac{1}{2} \mathrm{C}_{\mathrm{B}} \mathrm{V}^{2}\)
\(=\frac{1}{2} \times\left(3 \times 10^{-6}\right) \times(100)^{2}\)
= 1.5 x 10^{- 2} J
(ii) Q_A = C_AV = 2 x 10^-6 x 100
= 2 x 10^-4 C
QB = CBV = 3 x 10^-6 x 100
= 3 x 10^-4 C
Common Potential
\(=\mathrm{V}^{\prime}=\frac{\mathrm{Q}_{\mathrm{B}}-\mathrm{Q}_{\mathrm{A}}}{\mathrm{C}_{\mathrm{A}}+\mathrm{C}_{\mathrm{B}}}=\frac{(3-2) \times 10^{-4}}{(2+3) \times 10^{-6}}\)
= 20 V
(iii) Charge on capacitor A is
Q = CV' = 2 x 10^-6 x 120
= 4 x 10^{- 5} C
(iv) Here final electrostatic energy
\(=\frac{1}{2}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)\left(\mathrm{V}^{\prime}\right)^{2}\)
\(=\frac{1}{2}(2+3) \times 10^{-6} \times(20)^{2}\)
= 10^{- 3} J
Loss of energy = Initial electrostatic energy - final electrostatic energy
= (1.5 + 1.0) x 10^-2 - 10^{- 3}
= 2.4 x 10^{- 2} J

(i) R = 6.4 x 10⁶ m,
\(V=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{R}}\)      [ Earth is considered as a conducting sphere]
\(\therefore \quad C=\frac{Q}{\mathrm{~V}}=4 \pi \varepsilon_{0} R\)
\(\left[\because \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}\right]\)
\(=\frac{6.4 \times 10^{6}}{9 \times 10^{9}} \approx 711 \mu \mathrm{F}\)
(ii) Initially the potential of conductor B = V.

 When an uncharged conductor A is brought near to the conductor B, then the charges will induce on A. If -V₁ is the potential due to the negative induced charge at A, and V₂ is the potential due to the positive induced charge, then \(\left|-V_{1}\right|>\left|V_{2}\right|\) (as -ve charge is near to B).
Net potential on B = V-V₁ +V₂, which is less than V. But charge on B remains same.
(iii) E = 3 x 10⁶ V /m is the dielectric strength of air. If we further increase potential, the surrounding air will ionise and stored charge will leak away.
(iv) Principle: When an uncharged, earthed conductor is brought near to a charged conductor, then the potential of later decreases and its charge holding capacity increases.
The capacitance depends on:
(i) Geometrical configuration (shape, size and separation) of the system of two conductors.
(ii) Nature ofthe dielectric separating two conductors

(i) Capacitors are used to store electric charges and electric energy.
(ii) Both the protons will experience same force. Reason: F = qE; E = constant; q = +e (same)
(iii) \(\because\) V_D =.V_A > V_B = V_c
\(\therefore\) Gain in K.E. = q x P.D.
\( \therefore\) Gain in K.E. of proton released from point A will be more.
(iv) (a) \(W_{A \rightarrow B}=e\left(V_{B}-V_{A}\right)\)                  \(\left[\begin{array}{c} \because W=q \times \text { P.D. } \\ E=\frac{V}{d} \end{array}\right]\)
= e[E.2d - E.d]
= eE.d
(b) \(\because\) V_B = V_c
\(\therefore\) W_BC = 0
(c) W_CD = e[V_D - V_C]
= e[E.(d) - E(2d)]
= - eEd
(d) W_ABCD = W_AB + W_BC + W_CD + W_DA
= eEd + 0 - eEd + 0
= 0
(v) Electric field is conservative as work done along a closed path is zero.

(i) Given electric field on sphere S₁
E ₁ = 2 x 10⁴ N/C
\(\because \quad E_{1}=2 \times 10^{4} \mathrm{~N} / \mathrm{C}=\frac{k \mathrm{Q}}{r_{1}^{2}}\)
Now just inside outer sphere Electric field
\( E_{2}=\frac{k \mathrm{Q}}{r_{2}^{2}} \)
\(\frac{E_{2}}{E_{1}}=\frac{r_{1}^{2}}{r_{2}^{2}}\)
Here r₁ = 0.20 m, r ₂ = 0.40 m
\(E_{2}=2 \times 10^{4} \times\left(\frac{0.2}{0.4}\right)^{2}=0.5 \times 10^{4} \mathrm{~N} / \mathrm{c}\)
(ii) Electrostatic potential inside S₁
\(V_{1}=\frac{k \mathrm{Q}}{r_{1}}=r_{1} E_{1} \quad\left[\because \frac{E_{1}}{V_{1}}=\frac{1}{r_{1}}\right]\)
= 0.2 x 2 x 10⁴
= 0.4 x 10 ⁴ = 4 x 10 ³ V
(iii) If S₁ and S₂ are joined by a wire entire amount of energy stored in the system will get converted into heat.
Reason: Both the charges on spheres S₁ and S₂ will get neutralized and energy of the system will be dissipated as heat.

(i) (b): As, \(C_{1}=2 \mu F, C_{2}=4 \mu F\)
In series combination, the equivalent capacitance will be, \(C=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\left(\frac{2 \times 4}{2+4}\right) \mu \mathrm{F}=\frac{4}{3} \mu \mathrm{F}\)
Potential difference applied, V = 6 V
Energy stored in the system \(U=\frac{1}{2} C V^{2}\)
\(=\frac{1}{2} \times \frac{4}{3} \times 10^{-6} \times(6)^{2} \mathrm{~J}=24 \mu \mathrm{J}\)
(ii) (b): The energy stored in a capacitor is \(U=\frac{1}{2} C V^{2}=\frac{1}{2} \times\left(10 \times 10^{-6}\right)(10)^{2}=500 \mu \mathrm{J}\)
(iii) (b): When the gap between the plates is completely filled with dielectric of dielectric constant K, then potential is 
\(V=\frac{Q d}{A \varepsilon_{0} K}\)      ... (i)
and electric field is
\(E=\frac{Q}{A \varepsilon_{0} K}\)   ...(ii)
From equations (i) and (ii), both electric field and potential decrease.
(iv) (b): Work done = \(U_{f}-U_{i}=\frac{1}{2} \frac{q^{2}}{C_{f}}-\frac{1}{2} \frac{q^{2}}{C_{i}}\)
\(=\frac{q^{2}}{2}\left[\frac{1}{C_{f}}-\frac{1}{C_{i}}\right]=\frac{\left(5 \times 10^{-6}\right)^{2}}{2}\left[\frac{1}{2 \times 10^{-6}}-\frac{1}{5 \times 10^{-6}}\right]\)
\(=3.75 \times 10^{-6} \mathrm{~J}\)
(v) (b) : Here \(r=18 \mathrm{~cm}=18 \times 10^{-2} \mathrm{~m}, q=5 \times 10^{-6} \mathrm{C}\)
As \(C=4 \pi \varepsilon_{0} r=\frac{18 \times 10^{-2}}{9 \times 10^{9}}=2 \times 10^{-11} \mathrm{~F}\)
Energy of charged conductor is
\(U=\frac{q^{2}}{2 C}=\frac{\left(5 \times 10^{-6}\right)^{2} \mathrm{C}}{2 \times 2 \times 10^{-11} \mathrm{~F} \mid}=0.625 \mathrm{~J}\)

(i) (b): As the capacitor is isolated after charging, charge Q on it remains constant: Plate separation d Increases, capaci.tance decreases as \(C=\frac{\varepsilon_{0} A}{d}\) and hence, potential increases as \(V=\frac{Q}{C}\).
(ii) (c): In a parallel plate capacitor, the capacity of capacitor
 \(C=\frac{K \varepsilon_{Q} A}{d} \quad \text { i.e., } C \propto A\)
The capacity of capacitor increases if area of the plate increases.
(iii) (b): The magnitude of the electric field between the plates is E = \(\frac{\sigma}{2 \varepsilon_{0}}-\left(-\frac{\sigma}{2 \varepsilon_{0}}\right)=\frac{\sigma}{\varepsilon_{0}}\)
(IV) (b): As \(\frac{\varepsilon_{0} A}{d}=4 \pi \varepsilon_{0} R \text { or } \frac{\varepsilon_{0} \pi D^{2}}{4 d}=4 \pi \varepsilon_{0} R\)
\(\text { or } \quad d=\frac{D^{2}}{16 R}=\frac{(0.08)^{2}}{16 \times 0.10}=4 \times 10^{-3} \mathrm{~m}=4 \mathrm{~mm}\)
(v) (c): Here \(V=\frac{q_{1}-q_{2}}{2 C}\)
\(=\frac{2.0 \times 10^{-8}+1.0 \times 10^{-8}}{2 \times 1.2 \times 10^{-9}}=12.5 \mathrm{~V}\)

(i) (b): Here \(C-50 p F-50 \times 10^{-12} F, V=10^{4} V\)
\(R=\frac{1}{4 \pi \varepsilon_{0}} \cdot C=9 \times 10^{9} \mathrm{mF}^{-1} \times 50 \times 10^{-12} \mathrm{~F}\)
= 45 x 10^-2 m = 45 cm
(ii) (d): As \(q=C V=25 \times 10^{-12} \times 10^{5}=2.5 \mu \mathrm{C}\)
(iii) (c)
(iv) (c): As charge \(q=C V=\left(4 \pi \varepsilon_{0} R\right) V\)
\(\therefore\) q depends on both V and R.
(v) (c): 64 drops have formed a single mop of radius R. Volume oflarge sphere = 64 x Volume of small sphere
\(\therefore \frac{4}{3} \pi R^{3}=64 \frac{4}{3} \pi r^{3} \Rightarrow R=4 r \text { and } Q_{\text {total }}=64 q\)
\(C^{\prime}=4 \pi \varepsilon_{0} R \Rightarrow C^{\prime}=\left(4 \pi \varepsilon_{0}\right) \cdot 4 r \Rightarrow C^{\prime}=4 C\)

(i) (c): \(W=(\text { P.E. })_{\text {final }}-(\text { P.E. })_{\text {initial }}\)
\(=\frac{k e^{2}}{2}-\frac{k e^{2}}{1}=\frac{-k e^{2}}{2}\)
(ii) (b) : Potential at the centre of the square due to four equal charges q at four corners 
\(V=\frac{4 q}{4 \pi \varepsilon_{0}(a \sqrt{2}) / 2}=\frac{\sqrt{2} q}{\pi \varepsilon_{0} a}\)
\(W_{0 \rightarrow \infty}=-W_{\infty \rightarrow 0}=-(-q) V=\frac{\sqrt{2} q^{2}}{\pi \varepsilon_{0} a}\)
(iii) (c): Here, \(q=2 \mu \mathrm{C}=2 \times 10^{-6} \mathrm{C}, r_{A}=2 \mathrm{~m}, r_{B}=1 \mathrm{~m}\)
\(\therefore \ V_{A}-V_{B}=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{r_{A}}-\frac{1}{r_{B}}\right]\)
\(=2 \times 10^{-6} \times 9 \times 10^{9}\left[\frac{1}{2}-\frac{1}{1}\right] \mathrm{V}=-9 \times 10^{3} \mathrm{~V}\)
(iv)(b) : Required work done = Change in potential energy of the system
\(W=U_{f}-U_{i}=k \frac{q_{1} q_{2}}{r_{f}}-k \frac{q_{1} q_{2}}{r_{i}}=k q_{1} q_{2}\left[\frac{1}{r_{f}}-\frac{1}{r_{i}}\right]\)
\(\therefore \ W=\left(9 \times 10^{9}\right)\left(3 \times 10^{-9} \times 1 \times 10^{-9}\right)\) \(\times\left[\frac{1}{4 \times 10^{-2}}-\frac{1}{5 \times 10^{-2}}\right]\)
\(=27 \times 10^{-7} \times(0.05)=1.35 \times 10^{-7} \mathrm{~J}\)
(v) (b)

(i) (d)
(ii) (b)
(iii) (b)
(iv) (d): The electrical potential at any (J) point on circle of radius a due to charge Qa Q.at Its centre I.S V = \(V=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{a}\)              
It is an equipotential surface.
Hence, work done in carrying a charge q round the circle is zero.

(v) (d): Work done to move a unit charge along an equipotential surface from P to Q,
\(W=-\int_{P}^{Q} \vec{E} \cdot d \vec{l}\)
On equipotential surface \(\vec{E} \perp d \vec{l}\)
\(W=-\int_{P}^{Q} E(d l) \cos 90^{\circ}=0\)

(I) (c) : As \(\frac{-d V}{d r}=E_{r}\) the negative of the slope of V versus r curve represents the component of electric field along r. Slope of curve is zero only at point 3. Therefore, the electric field vector is zero at point 3.
(ii) (a) : Near positive charge, net potential is positive and near a negative charge, net potential is negative. Thus, charge Q₁ is positive and Q₂ is negative.
(iii) (b) : From the figure, it can be seen that net potential due to two charges is positive everywhere in the region left to charge Q₁. Therefore the magnitude of potential due to charge Q₁ is greater than due to Q_2.
(iv) (b)
(v) (c) : It can be seen that potential at the points both A and B are zero. When the charge is moved from A to B, work done by the electric field on the charge will be zero.

(i) (a) : As \(\Delta V=-E \Delta \psi=-\left(4.0 \times 10^{8} \mathrm{~V} / \mathrm{m}\right)(0.25 \mathrm{~m})\)= -10⁸V
(ii) (c) : As \(\Delta U=q_{0} \Delta V=\left(1.6 \times 10^{-19}\right) \times\left(-1.0 \times 10^{8} \mathrm{~V}\right)\)= \(-1.6 \times 10^{-11} \mathrm{~V}\)
(iii) (c) : Here, \(q_{1}=q_{2}=1.6 \times 10^{-19} \mathrm{C}, r=9 \times 10^{-15} \mathrm{~m}\)
\(U=\frac{9 \times 10^{9} \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{9 \times 10^{-15}}=2.56 \times 10^{-14} \mathrm{~J}\)
(iv) (a): Here, \(q_{1}=4 \mu \mathrm{C}, q_{2}=-3 \mu \mathrm{C}\)
r = 10 cm = 0.1 m 
Electrostatic potential energy,
\(U=\frac{1}{4 \pi \varepsilon_{o}} \frac{q_{1} q_{2}}{r}=9 \times 10^{9} \times \frac{4 \times 10^{-6} \times(-3) \times 10^{-6}}{0.1}=-1.1 \mathrm{~J}\)
(v) (a) : As proton moves in the direction of the electric field, then its potential energy decreases.

(i) (c)
(ii) (b)
(iii) (a): Since, E = 0 inside the conductor and has no tangential component on the surface, no work is done in moving a small test charge within the conductor and on its surface.
(iv) (b): The work done in bringing unit positive charge from infinity to a point which is at a distance x from the positive charge Q is defined as the potential at the given point due to the charge Q. Therefore
\(\phi=W\)
(v) (b): \(W_{\text {ext }}=q_{0} \Delta V\)
\(\left(W_{A B}\right)_{\mathrm{ext}}=q\left(V_{B}-V_{A}\right)\)
\(40 \mu \mathrm{J}=1 \mu \mathrm{C}\left(V_{B}-V_{A}\right)\)
\(V_{A}-V_{B}=-40 \mathrm{~V}\)