(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) 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.

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) (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): \(W_{\text {ext }}=q_{0} \Delta V\)
\(\left(W_{A B}\right)_{\mathrm{ext}}=q\left(V_{B}-V_{A}\right)\)
\(40 \mu \mathrm{J}=1 \mu \mathrm{C}\left(V_{B}-V_{A}\right)\)
\(V_{A}-V_{B}=-40 \mathrm{~V}\)