Electrostatics

Answer: (c)

1. Surface charge density of a hollow sphere: $\sigma = \dfrac{q}{4\pi R^2}$.
2. Same $q$, so $\sigma \propto \dfrac{1}{R^2}$.
3. $\dfrac{\sigma_1}{\sigma_2} = \dfrac{R_2^2}{R_1^2}$
4. So the answer is (c) $R_2^2 : R_1^2$.

Answer: (c)

1. In air $F = \dfrac{kq_1q_2}{r^2}$; in medium $F = \dfrac{kq_1q_2}{K r_1^2}$.
2. Equal forces: $K r_1^2 = r^2 \Rightarrow r_1 = \dfrac{r}{\sqrt{K}}$
3. So the answer is (c) $\dfrac{r}{\sqrt{K}}$.

Answer: (c)

1. $F \propto \dfrac{q_1q_2}{l^2}$.
2. $F' \propto \dfrac{2q_1q_2}{(l/2)^2} = \dfrac{8q_1q_2}{l^2} = 8F$
3. So the answer is (c) $n = 8$.

Answer: (a)

1. $\dfrac{kq^2}{d^2} = \dfrac{kq^2}{K D^2}$
2. $D^2 = \dfrac{d^2}{K} \Rightarrow D = \dfrac{d}{\sqrt{K}}$
3. So the answer is (a) $\dfrac{d}{\sqrt{K}}$.

Answer: (a)

1. In medium $F = \dfrac{kq_1q_2}{K d^2}$; in air at distance $d'$: $F = \dfrac{kq_1q_2}{d'^2}$.
2. $d'^2 = K d^2 \Rightarrow d' = d\sqrt{K}$
3. So the answer is (a) $d[K]^{1/2}$.

Answer: (b)

1. Vector from $q_1$ to $q_2$: $\vec r = (\hat i+\hat j+\hat k)-(2\hat i+3\hat j+3\hat k) = -\hat i-2\hat j-2\hat k$, $|\vec r| = 3\ \text{m}$.
2. $\vec F_2 = \dfrac{kq_1q_2}{r^3}\vec r = \dfrac{9\times10^9\times3\times10^{-6}\times(-4\times10^{-6})}{27}(-\hat i-2\hat j-2\hat k)$
3. $= -4\times10^{-3}(-\hat i-2\hat j-2\hat k) = (4\hat i+8\hat j+8\hat k)\times10^{-3}\ \text{N}$
4. So the answer is (b).

Answer: (b)

1. $q$ at the centre is in equilibrium for any value. Each end charge $Q$ must also be in equilibrium.
2. Let separation be $2r$: $\dfrac{kQ^2}{(2r)^2} + \dfrac{kQq}{r^2} = 0$
3. $\dfrac{Q}{4} + q = 0 \Rightarrow q = -\dfrac{Q}{4}$
4. So the answer is (b) $-\dfrac{Q}{4}$.

Answer: (a)

1. Net force on $+q$ zero: $\dfrac{ke q}{x^2} = \dfrac{k(4e)q}{(d-x)^2}$.
2. $\dfrac{d-x}{x} = 2 \Rightarrow x = \dfrac{d}{3}$
3. So the answer is (a) $\dfrac{d}{3}$ units.

Answer: (d)

1. For an end charge $+q$ (separation $2r$): $\dfrac{kq^2}{4r^2} + \dfrac{kqQ}{r^2} = 0$.
2. $Q = -\dfrac{q}{4}$
3. So the answer is (d) $\dfrac{-q}{4}$.

Answer: (d)

1. Let A, B carry $q$ each at distance $r$: $F = \dfrac{kq^2}{r^2} = 3\times10^{-5}\ \text{N}$.
2. C touches A: A and C each get $\dfrac{q}{2}$. C at midpoint, distance $\dfrac r2$ from each.
3. From B: $\dfrac{k(q/2)q}{(r/2)^2} = 2F$ (towards A); from A: $\dfrac{k(q/2)^2}{(r/2)^2} = F$ (towards B).
4. Net force on C $= 2F - F = F = 3\times10^{-5}\ \text{N}$
5. So the answer is (d) $x = 3$.

Answer: (a)

1. Coulomb force gives centripetal force: $\dfrac{kq_1q_2}{r^2} = m\omega^2 r = \dfrac{4\pi^2 m r}{T^2}$.
2. $T^2 = \dfrac{4\pi^2 m r^3}{kq_1q_2}$
3. So the answer is (a) $T = \left[\dfrac{4\pi^2 m r^3}{kq_1q_2}\right]^{1/2}$.

Answer: (b)

1. $F \propto q_1q_2$: initially $2\times6 = 12 \Rightarrow F = 12$ N, so $F = q_1q_2$ (in N) at this distance.
2. New charges: $2-4 = -2\ \text{C}$ and $6-4 = +2\ \text{C}$.
3. $F' = |-2\times2| = 4$ N; unlike charges, so attractive.
4. So the answer is (b) 4 N, attractive.

Answer: (d)

1. In air $F = \dfrac{kq_1q_2}{R^2}$; in a medium $F' = \dfrac{kq_1q_2}{KR_1^2}$.
2. Same force: $\dfrac{1}{R^2} = \dfrac{1}{KR_1^2} \Rightarrow R_1^2 = \dfrac{R^2}{K}$
3. So the answer is (d) $R_1 = \dfrac{R}{\sqrt{K}}$.

Answer: (b)

1. Force on $q$ from $Q$ (distance $d/3$): $\dfrac{kqQ}{(d/3)^2} = \dfrac{9kqQ}{d^2}$.
2. Force on $q$ from $4q$ (distance $2d/3$): $\dfrac{kq(4q)}{(2d/3)^2} = \dfrac{9kq^2}{d^2}$.
3. Both act along the same line; for zero net force $9kqQ = -9kq^2 \Rightarrow Q = -q$.
4. With the data as printed $Q = -q$, which is not among the options; the key gives (b) $-\dfrac{3q}{2}$.

Answer: (c)

1. Equilibrium: electric repulsion = gravitational attraction.
2. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{d^2} = \dfrac{Gm^2}{d^2} \Rightarrow \dfrac{q^2}{m^2} = 4\pi\varepsilon_0 G$
3. So the answer is (c) $\dfrac{q}{m} = \sqrt{4\pi\varepsilon_0 G}$.

Answer: (a)

1. The null point lies outside the charges, beyond the smaller charge $q_2$ (at $x > L$).
2. $\dfrac{k(8q)e}{x^2} = \dfrac{k(2q)e}{(x - L)^2} \Rightarrow \dfrac{x}{x - L} = 2$
3. $x = 2x - 2L \Rightarrow x = 2L$. So the answer is (a) $x = 2L$.

Answer: (c)

1. $F \propto \dfrac{q_1q_2}{l^2}$.
2. $F' \propto \dfrac{2q_1q_2}{(l/2)^2} = 8\dfrac{q_1q_2}{l^2}$
3. So the answer is (c) $n = 8$.

Answer: b

1. $\dfrac{F_e}{F_g} = \dfrac{kq_Aq_B}{Gm_Am_B}$ ($r$ cancels).
2. $= \dfrac{9\times10^9\times6.67\times10^{-19}\times9.6\times10^{-10}}{6.67\times10^{-11}\times19.2\times10^{-27}\times9\times10^{-27}}$
3. $= \dfrac{9\times9.6\times10^{-20}}{172.8\times10^{-65}} = 5\times10^{44}$
4. With the data as printed the ratio is $5\times10^{44}$, which cannot be written as $P\times10^{-13}$ with a sensible $P$.

Answer: (c)

1. From the figure, $q_1q_2$ is a side ($a$) and $q_1q_3$ is the diagonal ($\sqrt2 a$).
2. $F_{12} = \dfrac{kq^2}{a^2}$, $F_{13} = \dfrac{kq^2}{2a^2}$
3. $\dfrac{F_{13}}{F_{12}} = \dfrac{1}{2}$
4. So the answer is (c) $\dfrac12$.

Answer: (b)

1. Initially $F = \dfrac{kq^2}{r^2}$. After touching, the third ball and the touched ball each have $q/2$.
2. At the midpoint (distance $r/2$ from each): force from the untouched ball $= \dfrac{kq(q/2)}{(r/2)^2} = 2F$.
3. Force from the touched ball $= \dfrac{k(q/2)(q/2)}{(r/2)^2} = F$, in the opposite direction.
4. Net force $= 2F - F = F$. So the answer is (b) $F$.

Answer: (b)

1. Identical spheres share charge equally on contact.
2. C touches A: $A = C = \dfrac{q}{2}$. C touches B: $B = C = \dfrac{q/2 + q}{2} = \dfrac{3q}{4}$.
3. $F' = \dfrac{k(q/2)(3q/4)}{r^2} = \dfrac{3}{8}\dfrac{kq^2}{r^2} = \dfrac{3F}{8}$
4. So the answer is (b) $\dfrac{3F}{8}$.

Answer: (b)

1. After contact each sphere has $q = 2 \times 10^{-8}\text{ C}$.
2. $F = \dfrac{kq^2}{r^2} \Rightarrow r^2 = \dfrac{9 \times 10^9 \times 4 \times 10^{-16}}{9 \times 10^{-3}} = 4 \times 10^{-4}\text{ m}^2$
3. $r = 2 \times 10^{-2}\text{ m}$. So the answer is (b) 2 cm.

Answer: (b)

1. Force on $+q$ (at $l$) from $+3q$ (distance $l$): $\dfrac{3kq^2}{l^2}$, away from $+3q$.
2. Force on $+q$ from $Q$ (distance $l/2$): $\dfrac{kQq}{(l/2)^2} = \dfrac{4kQq}{l^2}$, same line.
3. Net zero: $3q + 4Q = 0 \Rightarrow Q = -\dfrac{3}{4}q$.
4. So $x = -\dfrac{3}{4}$, answer (b).

Answer: (a)

1. Proton beyond $q_2$ at distance $x$ from $q_1$: forces from $q_1$ (repulsion) and $q_2$ (attraction) must balance.
2. $\dfrac{6q}{x^2} = \dfrac{3q}{(x - L)^2} \Rightarrow \dfrac{x}{x - L} = \sqrt{2}$
3. $x(\sqrt{2} - 1) = \sqrt{2}L \Rightarrow x = \dfrac{\sqrt{2}}{\sqrt{2} - 1}L$
4. So the answer is (a).

Answer: (d)

1. Charge spreads over the outer surface: area $= 6a^2 = 6\times(0.05)^2 = 0.015\ \text{m}^2$.
2. $\sigma = \dfrac{q}{A} = \dfrac{6\times10^{-6}}{0.015} = 4\times10^{-4}\ \text{C m}^{-2}$
3. So the answer is (d) $0.4\times10^{-3}\ \text{C m}^{-2}$.

Answer: (b)

1. $\dfrac{F_e}{F_g} = \dfrac{ke^2}{G m_e m_p}$ (both $\propto 1/r^2$, so $r$ cancels).
2. $= \dfrac{9\times10^9\times(1.6\times10^{-19})^2}{6.67\times10^{-11}\times9.1\times10^{-31}\times1.67\times10^{-27}} \approx 2.3\times10^{39}$
3. So the answer is (b) $10^{39}$ (options mean $10^{19}, 10^{39}, \ldots$).

Answer: (a)

1. $\dfrac{kq^2}{x^2} = \dfrac{kq^2}{k_d\, y^2}$ with $k_d = k$ the dielectric constant.
2. $y^2 = \dfrac{x^2}{k} \Rightarrow \dfrac{y}{x} = \dfrac{1}{\sqrt{k}}$
3. So the answer is (a) $\dfrac{1}{\sqrt{k}}$.

Answer: (a)

1. $F' = \dfrac{1}{K}\dfrac{kq_1q_2}{(r/5)^2} = \dfrac{25}{K}F$
2. With $K = 5$: $F' = 5F$
3. So the answer is (a) $5F$.

Answer: (d)

1. $F \propto q_1q_2$ at fixed distance.
2. $F' = (1.1\,q_1)(0.9\,q_2)\,\dfrac{k}{r^2} = 0.99F = 99\text{ N}$
3. Change $= 99 - 100 = -1\text{ N}$. So the answer is (d) decreased by 1 N.

Answer: (c)

1. $F = \dfrac{k(4q)(4q)}{l^2} = \dfrac{16kq^2}{l^2}$
2. 25% of $4q$ is $q$. After transfer: $A = +3q$, $B = -4q + q = -3q$.
3. $F' = \dfrac{9kq^2}{l^2} = \dfrac{9}{16}F$
4. So the answer is (c) $\dfrac{9}{16}F$.

Answer: (b)

1. R touches P: $P = \dfrac{Q}{2}$, $R = \dfrac{Q}{2}$. R touches S: $S = \dfrac{Q/2 + Q}{2} = \dfrac{3Q}{4}$.
2. $F' = \dfrac{k(Q/2)(3Q/4)}{r^2} = \dfrac{3}{8}F$
3. $F' = \dfrac{3}{8} \times 16 = 6\text{ N}$. So the answer is (b) 6 N.

Answer: (d)

1. Let the side be $a$. Repulsion of $+Q$ by the other $+Q$ (distance $\sqrt{2}a$): $\dfrac{kQ^2}{2a^2}$ along the diagonal, outward.
2. Attraction by each $-q$ (distance $a$): $\dfrac{kQq}{a^2}$; the two are perpendicular, resultant $\dfrac{\sqrt{2}kQq}{a^2}$ along the diagonal, inward.
3. Balance: $\dfrac{kQ^2}{2a^2} = \dfrac{\sqrt{2}kQq}{a^2} \Rightarrow Q = 2\sqrt{2}\,q$.
4. So $\dfrac{+Q}{-q} = -2\sqrt{2}$, answer (d).

Answer: (d)

1. Force on $Q$ (at $x = 1$) from $3Q$ (distance 1): $\dfrac{3kQ^2}{1^2}$, along $+x$.
2. Force on $Q$ from $q$ (distance $\tfrac{2}{3}$): $\dfrac{kqQ}{(2/3)^2} = \dfrac{9kqQ}{4}$, along the same line.
3. Net zero: $3Q + \dfrac{9q}{4} = 0 \Rightarrow q = -\dfrac{4}{3}Q$.
4. So the answer is (d) $-\dfrac{4}{3}Q$.

Answer: 3

1. Same angle in air and liquid needs $K = \dfrac{\rho}{\rho - \sigma}$ (electric force $\div K$, weight reduced by buoyancy).
2. $K = \dfrac{1.5}{1.5 - 1} = 3$
3. Dielectric constant of water $= 3$.

Answer: 2

1. Same angle in air and liquid: $\dfrac{F}{mg} = \dfrac{F/K}{mg(1 - \sigma/\rho)} \Rightarrow K = \dfrac{\rho}{\rho - \sigma}$.
2. $K = \dfrac{1.4}{1.4 - 0.7} = 2$
3. Dielectric constant of the liquid $= 2$.

Answer: 3

1. Small element $R\,d\theta$ with charge $\lambda R\,d\theta$ ($\lambda = Q/2\pi R$) is pushed outward by the centre charge: $\dfrac{kq\lambda R\,d\theta}{R^2}$.
2. Balanced by tension: $2T\sin\dfrac{d\theta}{2} = T\,d\theta \Rightarrow T = \dfrac{kq\lambda}{R} = \dfrac{kqQ}{2\pi R^2}$
3. $T = \dfrac{9 \times 10^9 \times 30 \times 10^{-12} \times 2\pi}{2\pi \times 0.09} = \dfrac{0.27}{0.09}$
4. $T = 3\text{ N}$ (the cross-sectional area is not needed).

Answer: (c)

1. (a) False: electrostatic force is about $10^{36}$ times stronger.
2. (b) False: electrostatic force can be repulsive.
3. (c) True: both are central forces along the line joining the bodies.
4. (d) False: Coulomb force obeys $F \propto 1/r^2$.
5. So the answer is (c).