(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.

Only (c) is right; the rest cannot represent electrostatic field lines.
(a) is wrong because field lines must be normal to a conductor.
(b) is wrong because lines of force cannot start from a negative charge.
(d) is wrong because lines of force cannot intersect each other.
(e) is wrong because electrostatic field lines cannot form closed loops.

(i) The equivalent capacitance of C₁ and C₂ connected in series
\(\frac { 1 }{ C' } =\frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } \)
\(\Rightarrow C'=\frac { 6 }{ 2n } =3\mu F\)
∴ Charge, q' = C'V = (3μF)12 = 36μF
∴ Charge on each capadtor of C₁ and C₂ is 36 μC
∴ Charge on C₃,
q₃ = C₃V = (6μF) x 12 = 72μC
q₃ = 72μC
(ii) Equivalent capacitance of network,
\({ C }_{ eq }=\frac { { C }_{ 1 }{ C }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } } +{ C }_{ 3 }\)
\(=\frac { 6\times 6 }{ 6+6 } +6\)
= 3 + 6 = 9μF
C_eq = 9μF
(iii) Energy stored in the network of capacitors
U = U₁ + U₂ + U₃ = \(\frac { { q' }^{ 2 } }{ { 2C }_{ 1 } } +\frac { { q }'^{ 2 } }{ { 2C }_{ 2 } } +\frac { { q }^{ 2 } }{ { 2C }_{ 3 } } \)
∵ C₁ = C₂ = C₃ = 6μF
\(\therefore U=\frac { 1 }{ (12\mu F) } [{ q' }^{ 2 }+{ q' }^{ 2 }+{ q }^{ 2 }]\)
\(=\frac { 1 }{ (12\mu F) } [({ 36\mu C) }^{ 2 }+({ 36\mu C) }^{ 2 }+({ 72\mu C) }^{ 2 }]\)
\(U=648\mu J\)