"Electrostatics" — Practice

Question 1

A thin rod of length $L$ carries uniform linear charge density $\lambda$. Find the field at a point on its axis, a distance $a$ from the nearer end.

Accepted Answer

Take an element $dx$ at distance $x$ from the field point, with $x$ running from $a$ to $a+L$:

$E=\displaystyle\int_a^{a+L}\frac{k\lambda\,dx}{x^{2}}=k\lambda\left[-\frac1xight]_a^{a+L}$

$E=k\lambda\left(\frac1a-\frac{1}{a+L}ight)=\frac{k\lambda L}{a\left(a+Light)}$

For $a\gg L$ this becomes $kQ/a^{2}$ with $Q=\lambda L$, as it must.

Question 2

A quarter circle of radius $R$ carries total charge $Q$ uniformly. Find the magnitude of the field at the centre of curvature.

Accepted Answer

With $dq=\dfrac{2Q}{\pi}d	heta$ and $	heta$ from $0$ to $\pi/2$:

$E_x=\displaystyle\int_0^{\pi/2}\frac{2kQ}{\pi R^{2}}\cos	heta\,d	heta=\frac{2kQ}{\pi R^{2}}$

$E_y$ is the same by symmetry, so

$E=\sqrt2	imes\dfrac{2kQ}{\pi R^{2}}=\dfrac{2\sqrt2\,kQ}{\pi R^{2}}$

directed along the bisector of the quarter circle.

Question 3

In a region the potential is $V=3x^{2}y-y^{3}$ volts with distances in metres. Find the electric field at the point $(1,2)$.

Accepted Answer

$E_x=-\dfrac{\partial V}{\partial x}=-6xy$, and at $(1,2)$ this is $-12$ V m$^{-1}$.

$E_y=-\dfrac{\partial V}{\partial y}=-\left(3x^{2}-3y^{2}ight)=3y^{2}-3x^{2}$, and at $(1,2)$ this is $9$ V m$^{-1}$.

$\left|\vec Eight|=\sqrt{144+81}=15$ V m$^{-1}$

Question 4

A solid sphere of radius $R$ carries charge density $\rho=\rho_0\left(1-\dfrac{r}{R}\right)$. Find the field at $r<R$ and verify continuity at the surface.

Accepted Answer

$q_{enc}=\displaystyle\int_0^{r}ho_0\left(1-\frac{r'}{R}ight)4\pi r'^{2}dr'=4\piho_0\left(\frac{r^{3}}{3}-\frac{r^{4}}{4R}ight)$

$E=\dfrac{q_{enc}}{4\pi\varepsilon_0r^{2}}=\dfrac{ho_0}{\varepsilon_0}\left(\dfrac{r}{3}-\dfrac{r^{2}}{4R}ight)$

At $r=R$: $E=\dfrac{ho_0R}{12\varepsilon_0}$.

Total charge is $\piho_0R^{3}/3$, and $E_{out}$ at $r=R$ gives the same value, confirming continuity.

Question 5

A spherical cavity of radius $R/2$ is carved from a uniformly charged sphere of radius $R$ and density $\rho$, the cavity touching the centre and the surface. Find the field at the cavity's centre.

Accepted Answer

The cavity's centre is at $d=R/2$ from the sphere's centre.

$\vec E=\dfrac{ho\vec d}{3\varepsilon_0}$, so the magnitude is

$E=\dfrac{ho R}{6\varepsilon_0}$

directed from the sphere's centre towards the cavity's centre. Crucially, the same value holds at **every** point of the cavity, not just at its centre.

Question 6

Three concentric conducting shells of radii $a$, $2a$ and $3a$ carry charges $q$, $-2q$ and $3q$. The middle shell is now earthed. Find the charge that ends up on it.

Accepted Answer

Let the middle shell carry $Q'$ after earthing. Its potential is

$V=\dfrac{kq}{2a}+\dfrac{kQ'}{2a}+\dfrac{k\left(3qight)}{3a}=0$

Multiplying by $\dfrac{2a}{k}$: $q+Q'+2q=0$

$Q'=-3q$

The outermost shell contributes through its own radius, which is why it enters with the factor $2/3$.

Question 7

A charge $q$ sits at the centre of an uncharged conducting spherical shell of inner radius $a$ and outer radius $b$. Find the potential at a point inside the cavity, at distance $r<a$ from the centre.

Accepted Answer

Induced charges are $-q$ spread over $r=a$ and $+q$ over $r=b$.

Inside a uniformly charged shell the potential is constant at $kq'/R$, so each induced surface contributes its own value at any interior point:

$V(r)=\frac{kq}{r}-\frac{kq}{a}+\frac{kq}{b}$

$V(r)=kq\left(\frac1r-\frac1a+\frac1bight)$

At $r=a$ this reduces to $kq/b$, which is correctly the constant potential of the conductor itself.

Question 8

A charge $q$ sits at distance $d$ from a large grounded plane. Find the total induced charge on the plane and the electric field at the point of the plane directly below the charge.

Accepted Answer

The induced surface density is $\sigma(s)=-\dfrac{qd}{2\pi\left(s^{2}+d^{2}ight)^{3/2}}$, and integrating over the plane gives exactly $-q$.

Directly below the charge, $s=0$:

$\sigma(0)=-\dfrac{q}{2\pi d^{2}}$

$E=\dfrac{\sigma}{\varepsilon_0}=-\dfrac{q}{2\pi\varepsilon_0d^{2}}$, directed into the surface, which is also $2kq/d^{2}$ — the sum of the real and image charges' fields there.

Question 9

Find the self-energy of a uniformly charged solid sphere of radius $R$ and total charge $Q$ by assembling it shell by shell.

Accepted Answer

At intermediate radius $r$ the built-up charge is $q=Q\dfrac{r^{3}}{R^{3}}$, and the added shell carries $dq=Q\dfrac{3r^{2}dr}{R^{3}}$.

$dU=\dfrac{k\,q\,dq}{r}=\dfrac{3kQ^{2}r^{4}dr}{R^{6}}$

$U=\displaystyle\int_0^{R}\frac{3kQ^{2}r^{4}}{R^{6}}dr=\frac{3kQ^{2}}{5R}$

Integrating $	frac12\varepsilon_0E^{2}$ over all space, inside and out, gives the same value.

Question 10

A dipole of moment $p$ sits at the origin pointing along $x$. A second identical dipole is placed at distance $r$ along the $y$ axis, also pointing along $x$. Find the potential energy of the pair.

Accepted Answer

On the equatorial line the field of the first dipole is $E=\dfrac{kp}{r^{3}}$, directed **opposite** to $\vec p$, that is along $-x$.

The second dipole points along $+x$, so $\vec p\cdot\vec E=-\dfrac{kp^{2}}{r^{3}}$.

$U=-\vec p\cdot\vec E=+\dfrac{kp^{2}}{r^{3}}$

The energy is positive, so parallel side-by-side dipoles repel — the opposite of the collinear arrangement, which attracts.

Question 11

A parallel-plate capacitor of area $A$ and gap $d$ is connected to a battery of emf $V$. A dielectric slab of constant $K$ and thickness $d$ is inserted to fill the gap completely. Find the change in stored energy and the work done by the battery.

Accepted Answer

$C_0=\dfrac{\varepsilon_0A}{d}$ becomes $KC_0$.

$\Delta U=	frac12\left(KC_0ight)V^{2}-	frac12C_0V^{2}=	frac12C_0V^{2}\left(K-1ight)$

Charge rises by $\Delta Q=C_0V\left(K-1ight)$, so the battery supplies

$W_{battery}=V\Delta Q=C_0V^{2}\left(K-1ight)$

The battery supplies **twice** the increase in stored energy; the other half is the work done pulling the slab in, later dissipated.

Question 12

Two capacitors of $3\ \mu$F and $6\ \mu$F are charged to $100$ V and $50$ V respectively, then connected positive plate to positive plate. Find the common potential and the energy lost.

Accepted Answer

$V=\dfrac{C_1V_1+C_2V_2}{C_1+C_2}=\dfrac{3	imes100+6	imes50}{9}=\dfrac{600}{9}=66.7$ V

$\Delta U=\dfrac{C_1C_2\left(V_1-V_2ight)^{2}}{2\left(C_1+C_2ight)}=\dfrac{18	imes2500}{18}	imes10^{-6}$

$\Delta U=2.5	imes10^{-3}$ J

The loss occurs however low the wire resistance is; lowering it only makes the dissipation faster.

Practice — "Electrostatics"

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