Electrostatics Questions and Answers

Answer: E

The field is \(2 \times 10^{5}\) V m^-1.

For a uniform field between conductors, \(E = \frac{V}{d}\), with \(d\) in metres:

\[ \begin{aligned} E &= \frac{4 \times 10^{3}\ \text{V}}{2 \times 10^{-2}\ \text{m}} \\ &= 2 \times 10^{5}\ \text{V m}^{-1} \end{aligned} \]

\(2 \times 10^{3}\) uses \(d = 2\) m, forgetting the centimetres. \(8 \times 10^{3}\) and \(8 \times 10^{5}\) multiply by the distance instead of dividing. \(4 \times 10^{5}\) divides by 1 cm instead of 2.

Answer: A

The energy density is proportional to \(E^{2}\).

The energy stored per unit volume in an electric field is:

\[ u = \frac{1}{2}\varepsilon_{0}E^{2} \]

It follows from the capacitor energy \(\frac{1}{2}CV^{2}\) with \(C = \frac{\varepsilon_{0}A}{d}\) and \(V = Ed\), divided by the volume \(Ad\). Doubling the field quadruples the stored energy density.

\(E\), \(\sqrt{E}\) and \(E^{3}\) have the wrong power, and \(E^{-2}\) would make a stronger field store less energy.

Answer: C

Polarization is the dipole moment per unit volume.

When a dielectric is placed in an electric field, its molecules acquire or align dipole moments. The polarization \(\vec{P}\) is the net dipole moment per unit volume; for a linear dielectric \(\vec{P} = \chi_{e}\varepsilon_{0}\vec{E}\).

Its SI unit is therefore C m / m³ = C m^-2. Dipole moment per unit area, length, charge or field is not the definition.

Answer: E

The energy density is \(2\varepsilon_{0}\).

The energy stored per unit volume in an electric field is

\[ \begin{aligned} u &= \frac{1}{2}\varepsilon_{0}E^{2} \\ &= \frac{1}{2}\varepsilon_{0}(2)^{2} = 2\varepsilon_{0} \end{aligned} \]

in J m^-3. \(4\varepsilon_{0}\) forgets the \(\frac{1}{2}\), and \(\frac{\varepsilon_{0}}{2}\) forgets to square \(E\).

The capacitance C does not appear: energy density depends only on the field at the point (and the medium), not on the capacitor.

Answer: A

The ratio is 2 : 1.

Joined by a wire, the spheres reach the same potential: \(\frac{kq_{1}}{r_{1}} = \frac{kq_{2}}{r_{2}}\), so \(q \propto r\).

\[ \begin{aligned} E &= \frac{kq}{r^{2}} \propto \frac{r}{r^{2}} = \frac{1}{r} \\ \frac{E_{1}}{E_{2}} &= \frac{r_{2}}{r_{1}} = \frac{2}{1} \end{aligned} \]

The smaller sphere has the stronger surface field, which is why charge density is greatest at sharp points of a conductor. The option 1 : 2 is the ratio of the charges, not of the fields.

Answer: C

The work done is 20 eV.

Work done by a uniform field on a charge moving along it is \(W = qEd\). For a proton, \(q = e\):

\[ \begin{aligned} W &= e \times 4\ \text{N C}^{-1} \times 5\ \text{m} \\ &= 20\ e\text{V} \end{aligned} \]

Expressing the work in electronvolts avoids the value of \(e\): it is \(20 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-18}\) J. The proton moves along the field, so the work is positive.

Answer: D

The charge is 1 millicoulomb.

Torque on a dipole is \(\tau = pE\sin\theta\), with dipole moment \(p = q \times 2a\) (the length, here 2 cm).

\[ \begin{aligned} 3 &= q(0.02)(3 \times 10^{5})\sin 30^{\circ} \\ 3 &= q \times 3000 \\ q &= 10^{-3}\ \text{C} = 1\ \text{mC} \end{aligned} \]

Forgetting \(\sin 30^{\circ} = \frac{1}{2}\) gives 0.5 mC, and using 2 cm as half-length gives 0.5 mC as well; neither is offered.

Answer: B

The effective capacitance is \(\frac{kC}{1 + k}\).

Filling one capacitor with the dielectric raises its capacitance to \(kC\). The two are in series:

\[ \begin{aligned} \frac{1}{C_{eff}} &= \frac{1}{C} + \frac{1}{kC} = \frac{k + 1}{kC} \\ C_{eff} &= \frac{kC}{1 + k} \end{aligned} \]

Check: with \(k = 1\) (no dielectric) this gives \(\frac{C}{2}\), the usual result for two equal capacitors in series. The option \((k + 1)C\) is the parallel combination, and \(\frac{2kC}{1 + k}\) is double the correct value.

Answer: D

The charges are +ne and -(m+n)e.

The first sphere is neutral and loses n electrons, so it is left with a positive charge of +ne, where e is the magnitude of the electronic charge.

For the second sphere, the question's 'having m electrons' has to be read as an excess of m electrons, a charge of -me, since a body's own electrons are otherwise balanced by its protons. Gaining n more electrons makes its charge -(m+n)e. The options with -ne for the first sphere get the sign wrong, since losing electrons leaves a body positive, and those with (m-n) subtract the transferred electrons instead of adding them.

Answer: B

The capacitance is still 100 µF.

Capacitance depends only on the capacitor's construction: the area of the plates, the distance between them and the dielectric between them (C = εA/d for a parallel-plate capacitor). It does not depend on the charge stored. Removing half the charge halves the voltage across the capacitor, since V = Q/C, but C itself is unchanged.

50 µF wrongly scales C with the charge, and 200 µF scales it inversely. 5 µF and 32 µF have no connection with the data.