Electrostatics Questions and Answers

Answer: C

The potential at C is 2K.

Potential is a scalar, so the contributions simply add. C is at distance R from both A and B:

\[ \begin{aligned} V &= \frac{Q}{4\pi\epsilon_{0}R} + \frac{Q}{4\pi\epsilon_{0}R} \\ &= 2K \end{aligned} \]

The options with \(\sqrt{2}\), \(\sqrt{3}\) or \(\sqrt{5}\) treat potential as a vector and combine the contributions by geometry; that is how the electric field at C would be found (giving \(\sqrt{3}\) times one field), not the potential.

Answer: E

The effective capacitance is 300 pF.

Two 100 pF capacitors in parallel give 100 + 100 = 200 pF. Two 200 pF capacitors in series give \(\frac{200 \times 200}{400} = 100\) pF.

Connecting one combination across the other places them in parallel, so the capacitances add:

\[ C = 200 + 100 = 300\ \text{pF} \]

200 pF and 100 pF are the two combinations on their own. Putting them in series instead would give \(\frac{200 \times 100}{300} \approx 66.7\) pF.

Answer: B

The work done is 6 J.

Work to rotate a dipole from \(\theta_{1}\) to \(\theta_{2}\) in a uniform field is \(W = pE(\cos\theta_{1} - \cos\theta_{2})\).

\[ \begin{aligned} W &= 12 \times 10^{-6} \times 10^{6} \times (1 - \tfrac{1}{2}) \\ &= 12 \times \tfrac{1}{2} = 6\ \text{J} \end{aligned} \]

12 J is pE, the work for a rotation from 0° to 90°. Using \(\sin 60^{\circ}\) gives about 10.4 J, which is not offered.

Answer: A

The separation is 2.0 cm.

In a uniform field, the potential difference is \(V = Ed\), so \(d = \frac{V}{E}\).

\[ \begin{aligned} d &= \frac{16}{800} = 0.02 \text{ m} \\ &= 2.0 \text{ cm} \end{aligned} \]

The unit N C^-1 is the same as V m^-1, so dividing volts by volts per metre gives metres. The only step to watch is the conversion from metres to centimetres; the other values do not satisfy \(16 = 800 \times d\).

Answer: D

The dielectric constant is 3.

Filling a capacitor with a dielectric multiplies its capacitance by the dielectric constant K. The capacitance 'changes by 2C', meaning it increases by 2C, from C to 3C.

\[ K = \frac{C + 2C}{C} = 3 \]

2 is the trap of reading 'changes by 2C' as 'becomes 2C'. A dielectric constant of 1 would mean no change at all.

Answer: E

The field is 0.72 NC^-1.

The charge is \(q = 5 \times 10^{6} \times 1.6 \times 10^{-19} = 8 \times 10^{-13}\) C, and it resides on the surface. At the surface the field is as if the charge were at the centre:

\[ \begin{aligned} E &= \frac{kq}{r^{2}} = \frac{9 \times 10^{9} \times 8 \times 10^{-13}}{(0.1)^{2}} \\ &= \frac{7.2 \times 10^{-3}}{0.01} = 0.72\ \text{N C}^{-1} \end{aligned} \]

1.44 doubles the result, and 0.36 halves it. The radius must be in metres.

Answer: B

The charge is 1.8 nC.

The stored energy is \(U = \frac{Q^{2}}{2C}\), so

\[ \begin{aligned} Q^{2} &= 2CU = 2 \times 9 \times 10^{-12} \times 18 \times 10^{-8} \\ &= 3.24 \times 10^{-18} \\ Q &= 1.8 \times 10^{-9}\ \text{C} = 1.8\ \text{nC} \end{aligned} \]

Using \(U = \frac{Q^{2}}{C}\) without the factor 2 gives about 1.27 nC, and 3.6 nC doubles the answer.

Answer: D

The field has magnitude 10 Vm^-1.

The electric field is the negative gradient of potential:

\[ \begin{aligned} E &= -\frac{dV}{dx} = -(6x + 4) \\ E(1) &= -10\ \text{V m}^{-1} \end{aligned} \]

Its magnitude is 10 V m^-1, directed towards negative \(x\). 6 and 4 take only one of the two terms of the derivative, and 8 and 12 do not come from \(6x + 4\) at \(x = 1\).

Answer: E

The correct statement is In a non-polar molecule, centres of positive and negative charges coincide.

That is the definition of a non-polar molecule: with coincident charge centres it has no permanent dipole moment.

The others are false. In a polar molecule the centres do not coincide, giving a permanent dipole. Water is the standard example of a polar molecule because of its bent shape. Dielectrics are insulators, not conductors. The statement about air is loose at best: its main gases, nitrogen and oxygen, are non-polar, but air also contains polar molecules such as water vapour, so it is not true of air's molecules as a whole.

Answer: E

The new capacitance is Ck.

For a parallel plate capacitor \(C = \frac{\varepsilon_0 A}{d}\) in air. Filling the space with a dielectric of constant k multiplies the permittivity by k, so \(C' = \frac{k\varepsilon_0 A}{d} = kC\).

The dielectric is polarised and reduces the field between the plates for a given charge, so the plates hold more charge at the same voltage. The capacitance therefore rises; it does not fall to \(\frac{C}{k}\), and it grows in proportion to k, not to \(k^{2}\).