Electrostatics and Capacitors

(c) $4\sqrt{35}$ N

$\vec E=-\nabla V$. Here $E_x=-(6-8y)=2$, $E_y=-(-8x-8+6z)=10$, $E_z=-(6y)=-6$ at $(1,1,1)$. $|E|=\sqrt{4+100+36}=\sqrt{140}=2\sqrt{35}$ V/m. Force $F=qE=2\times2\sqrt{35}=4\sqrt{35}$ N. Forgetting the factor $q=2$ would give $2\sqrt{35}$.

(d) $\frac{CV^2}{2d}$

Energy stored is $U=\frac12CV^2$. The attractive force is found from $F=\frac{dU}{dx}$ at constant $V$ with $C=\frac{\varepsilon_0A}{x}$. This gives $F=\frac12V^2\left|\frac{dC}{dx}\right|=\frac12V^2\frac{C}{d}=\frac{CV^2}{2d}$. Option (a) misses the factor $\frac12$.

(d) $-(6\hat i+5\hat j+2\hat k)$

The field is the negative gradient of potential: $\vec E = -\left(\frac{\partial V}{\partial x}\hat i+\frac{\partial V}{\partial y}\hat j+\frac{\partial V}{\partial z}\hat k\right)$. Here $\partial V/\partial x = 6y$, $\partial V/\partial y = 6x-1+2z$ and $\partial V/\partial z = 2y$. At $(1,1,0)$ these are $6$, $5$ and $2$, so $\vec E = -(6\hat i+5\hat j+2\hat k)$.

(d) $v\propto x^{-1/2}$

In equilibrium (small angle) the electric force balances the horizontal component of tension: $\dfrac{q^2}{4\pi\varepsilon_0x^2}\propto \dfrac{x}{2l}mg$, so $q^2\propto x^3$ and $q\propto x^{3/2}$. The charge leaks at a constant rate: $\dfrac{dq}{dt}\propto x^{1/2}\dfrac{dx}{dt}=$ constant. Thus $v=\dfrac{dx}{dt}\propto x^{-1/2}$.

(a) 80%

Initially $U_i=\tfrac12CV^2=\tfrac12(2)V^2=V^2$ (in $\mu$J). The charge $2V$ is shared by $2+8=10\ \mu$F, so $U_f=\dfrac{(2V)^2}{2\times10}=0.2V^2$. Energy lost is $0.8V^2$, which is 80% of the initial energy (dissipated in the connecting wires/spark). Energy is not conserved in charge sharing, so (b) is wrong.

(a) decreases by a factor of 2

Initial energy $U=\tfrac{q^2}{2C}$. After connecting an identical uncharged capacitor in parallel, charge is shared (total $q$) and capacitance doubles, so $U'=\tfrac{q^2}{2(2C)}=U/2$.

The missing half is lost as heat/radiation in the connecting wires; energy is not conserved in the capacitor system alone.

(b) $10^{-37}$ C

Each hydrogen atom has a net charge $\Delta e$ (proton and electron charges fail to cancel exactly). Balancing electrostatic repulsion with gravitational attraction: $\dfrac{(\Delta e)^2}{4\pi\varepsilon_0d^2}=\dfrac{Gm_h^2}{d^2}$.

$\Delta e=m_h\sqrt{4\pi\varepsilon_0G}=1.67\times10^{-27}\sqrt{6.67\times10^{-11}/(9\times10^9)}\approx4.5\times10^{-38}$ C, of order $10^{-37}$ C.

(b) independent of the distance between the plates

For an isolated capacitor the charge $Q$ is fixed. The field of one plate is $\sigma/2\epsilon_0$, so $F = Q\cdot \dfrac{Q}{2\epsilon_0 A} = \dfrac{Q^2}{2\epsilon_0 A}$. There is no $d$ in the result: the field between the plates stays constant as they are moved.

(b) smaller

From rest, $h = \tfrac12 a t^2$, so $t = \sqrt{2h/a}$; a larger acceleration gives a shorter time. Electron: $a_e = g + eE/m_e$ (field up, force on electron down). Proton after reversal: $a_p = g + eE/m_p$. Since $m_e \ll m_p$, $a_e \gg a_p$, so the electron takes less time.

(b) zero as $r$ increases for $r<R$, decreases as $r$ increases for $r>R$

Inside a uniformly charged hollow sphere, Gauss's law gives zero enclosed charge, so $E=0$ for $r<R$. Outside, the sphere acts as a point charge at the centre: $E=\dfrac{Q}{4\pi\varepsilon_0r^2}$, which decreases with $r$.

(c) $\dfrac{\lambda}{\pi\varepsilon_0R}$ N/C

Each line gives $E=\dfrac{\lambda}{2\pi\varepsilon_0R}$ at the mid-point (distance $R$ from each). The field of $+\lambda$ points away from it and that of $-\lambda$ points toward the other line, so both point the same way and add: $E=\dfrac{\lambda}{\pi\varepsilon_0R}$. Zero would hold only for like charges.

(b) $\dfrac{9F}{16}$

Initially $F=\dfrac{kQ^2}{r^2}$. After moving $0.25Q$ from A to B, A has $0.75Q$ and B has $-Q+0.25Q=-0.75Q$. $F'=\dfrac{k(0.75Q)^2}{r^2}=\dfrac{9}{16}F$. The force falls since the product of charges decreases.

(d) 200 V

Potential of a short dipole: $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$. $V=\dfrac{9\times10^9\times16\times10^{-9}\times0.5}{0.36}=\dfrac{72}{0.36}=200$ V. Zero would be for $\theta=90^\circ$.

(a) $0.44\times10^{-10}\ \mathrm{C^2\,N^{-1}\,m^{-2}}$

The dielectric constant is $K=\dfrac{C}{C_0}=\dfrac{30}{6}=5$. Permittivity $\varepsilon=K\varepsilon_0=5\times8.85\times10^{-12}=4.4\times10^{-11}=0.44\times10^{-10}$. Option (b) is $K$, which is dimensionless.

(b) Having a permanent electric dipole moment.

Polar molecules (such as $H_2O$, HCl) have centres of positive and negative charge that do not coincide, so they carry a permanent dipole moment even without a field. Option (d) describes non-polar molecules, which acquire an induced dipole moment only when an electric field displaces their charges.

(d) $\frac{R_2}{R_1}$

Connected by a wire, the two spheres reach the same potential: $\frac{kq_1}{R_1}=\frac{kq_2}{R_2}$, so $q\propto R$. Then $\sigma=\frac{q}{4\pi R^2}\propto\frac{1}{R}$, giving $\frac{\sigma_1}{\sigma_2}=\frac{R_2}{R_1}$. The smaller sphere has the larger charge density.

(a) $\frac{1}{2}\varepsilon_0E^2Ad$

The energy density of an electric field is $\frac{1}{2}\varepsilon_0E^2$. The field occupies volume $Ad$, so $U=\frac{1}{2}\varepsilon_0E^2Ad$. Option (c) is only the energy density, not the total energy.

(b) 1980 V

Let each drop have radius $r$ and charge $q$. Volume conservation: $R^3=27r^3$, so $R=3r$. Charge $Q=27q$. $V'=\frac{kQ}{R}=\frac{27kq}{3r}=9\cdot\frac{kq}{r}=9\times220=1980$ V.

(b) More on smaller sphere

For a conducting sphere, $V = \dfrac{kq}{R}$. With equal charge $q$, the potential is inversely proportional to the radius, so the smaller sphere ($R_2$) is at the higher potential. Potential depends on charge and radius only, not on the material, so (d) is wrong.

(c) $90^\circ$

The electric field is always perpendicular to an equipotential surface. If it had a component along the surface, work would be done moving a charge along it, contradicting constant potential.

(b) $\dfrac{1}{R^3}$

Two equal and opposite charges separated by $L$ form a dipole. At distances $R\gg L$ the field of a dipole falls as $1/R^3$ (on axis $2kp/R^3$). The $1/R^2$ law applies only to a single charge, because the fields of the two charges nearly cancel.

(c) $2.25\times10^{-6}$ J

Initial energy $=\tfrac12CV^2 = 0.5\times900\times10^{-12}\times10^4 = 4.5\times10^{-6}$ J. Charge $Q$ is conserved and shared over $2C$, so the common voltage is $V/2$ and the final energy is $\tfrac12(2C)(V/2)^2 = \tfrac12\cdot\tfrac12CV^2 = 2.25\times10^{-6}$ J. Half the energy is lost as heat, so 4.5 is wrong.

(d) 2 mC

Torque $\tau = pE\sin\theta = qLE\sin\theta$. So $q = \dfrac{4}{0.02\times2\times10^5\times0.5} = \dfrac{4}{2000} = 2\times10^{-3}$ C $= 2$ mC.

(a) the number of flux lines entering the surface must be equal to the number of flux lines leaving it.

Zero net flux means that the flux entering equals the flux leaving, so the net enclosed charge is zero. It does not mean $E$ is constant or uniform, and charges may still lie inside (as +q and -q) or outside the surface.

(b) A is true but R is false

Axial potential is $V=\pm\frac{p}{4\pi\varepsilon_0r^2}=\pm\frac{9\times10^9\times4\times10^{-6}}{4}=\pm9\times10^3$ V, so A is true. The formula in R has an extra factor 2 (that factor belongs to the axial field, not the potential), so R is false.

(a) A, C and E only

The capacitor stays connected to the battery, so $V$ is constant. Bringing the plates closer raises $C=\frac{\varepsilon_0A}{d}$ (C is true). Then $Q=CV$ increases (A is true) and $U=\frac12CV^2$ increases, so B is false. $Q/V=C$ changes, so D is false. $QV$ increases because $Q$ does (E is true).

(d) 1.0 J

Potential energy is $U=-pE\cos\theta$, so $\Delta U=pE(\cos0^\circ-\cos60^\circ)=pE\left(1-\frac12\right)$. $pE=5\times10^{-6}\times4\times10^{5}=2$ J, so $\Delta U=1.0$ J. Option (a) 1.2 J does not follow from any consistent working.

(b) $\frac{3F}{8}$

Touching A with the neutral sphere leaves each with $\frac q2$. Touching B (charge $q$) then gives each $\frac{q/2+q}{2}=\frac{3q}{4}$. New force $=k\frac{(q/2)(3q/4)}{r^2}=\frac38\cdot\frac{kq^2}{r^2}=\frac{3F}{8}$. The common error is to assume A and B both end at $\frac q2$.

(c) 2.66

The air gap is $d-\frac{3d}{8}-\frac d2=\frac d8$. The layers are in series, so $C=\frac{\varepsilon_0A}{\frac d8+\frac{3d}{8K_1}+\frac{d}{2K_2}}=2\,\frac{\varepsilon_0A}{d}$. This gives $\frac18+\frac{3}{8K_1}+\frac{1}{2K_2}=\frac12$, i.e. $\frac{3}{8K_1}+\frac{1}{2K_2}=\frac38$. With $K_1=1.25K_2$: $\frac{0.3}{K_2}+\frac{0.5}{K_2}=\frac38$, so $K_2=2.13$ and $K_1=1.25\times2.13\approx2.66$.