Electrostatics Questions and Answers

Answer: D

The maximum force is between 0.5 C and 0.5 C.

Every pair adds up to 1 C, and at the same separation Coulomb's force is proportional to the product \(q_{1}q_{2}\). For a fixed sum, the product is largest when the two parts are equal.

\[ \begin{aligned} 0.5 \times 0.5 &= 0.25 \\ 0.4 \times 0.6 = 0.24,\ 0.3 \times 0.7 &= 0.21 \\ 0.2 \times 0.8 = 0.16,\ 0.1 \times 0.9 &= 0.09 \end{aligned} \]

So the equal split gives the largest product and hence the largest force; the more unequal the split, the smaller the force.

Answer: C

The potential at Q is 2.5 V.

The work done in moving a charge q from P to Q equals \(q(V_{Q} - V_{P})\).

\[ \begin{aligned} 10 &= 5(V_{Q} - 0.5) \\ V_{Q} - 0.5 &= 2 \Rightarrow V_{Q} = 2.5\ \text{V} \end{aligned} \]

2.0 V is the potential difference, not the potential at Q. 1.5 V subtracts the 0.5 V instead of adding it, which would mean work was done by the field rather than against it.

Answer: A

Each capacitor is 2 μF.

For n equal capacitors C, the series combination is \(\frac{C}{n}\) and the parallel one is nC.

\[ \begin{aligned} \frac{C}{n} \times nC &= C^{2} = 0.4 \times 10 = 4 \\ C &= 2\ \mu\text{F},\quad n = 5 \end{aligned} \]

Check: 2/5 = 0.4 and 5 x 2 = 10. 4 μF is \(C^{2}\), and 5 μF is the number of capacitors n.

Answer: D

The field inside is zero.

By Gauss's law, a spherical surface drawn inside the shell encloses no charge, so the flux through it is zero. By symmetry the field is the same in magnitude everywhere on that surface and radial, so it must be zero at every point inside.

There is no charge within the shell for the field to be proportional to, the \(\frac{1}{R^{2}}\) and 'same as outside' options describe the field outside, and nothing makes it maximum at the centre.

Answer: E

The torque is P × E.

The two charges feel equal and opposite forces qE, which form a couple. Its moment is \(qE \times 2a\sin\theta = pE\sin\theta\), and its direction is given by the cross product, so \(\vec{\tau} = \vec{p} \times \vec{E}\).

qE and −qE are the forces on the individual charges, not the torque. The torque is zero only when the dipole is aligned with the field, and P·E (with a minus sign) is the potential energy of the dipole, not its torque.

Answer: C

The field of a positive charge points radially outwards away from the charge.

The field at any point is the force on a small positive test charge placed there. A positive charge repels it along the line joining them, so the field lines start on the charge and point radially outwards in all directions.

Radially inward fields belong to negative charges. Circular field lines are characteristic of the magnetic field around a current, and the field is not along a single fixed line.

Answer: A

The new force is 25 N.

Coulomb's force is proportional to the product of the charges at a fixed separation. On contact, identical conductors share the total charge equally: \(\frac{8 + 2}{2} = 5\ \mu\text{C}\) each.

\[ \begin{aligned} \frac{F'}{F} &= \frac{5 \times 5}{8 \times 2} = \frac{25}{16} \\ F' &= 16 \times \frac{25}{16} = 25\ \text{N} \end{aligned} \]

The force increases because, for a fixed total, the product is largest when the charges are equal. 20 N and 15 N are slips in the ratio, and 16 N (not offered) would assume nothing changes.

Answer: B

The potential difference is 4 V.

Potential difference is the work done per unit charge in moving a charge between the two points:

\[ \begin{aligned} V &= \frac{W}{q} = \frac{12\ \text{J}}{3\ \text{C}} \\ &= 4\ \text{V} \end{aligned} \]

One volt is one joule per coulomb. 36 V multiplies the work by the charge instead of dividing, and the other values do not come from W/q.

Answer: C

The torque is \(\frac{PE}{2}\).

A dipole in a uniform field experiences a torque \(\tau = PE\sin\theta\), where \(\theta\) is the angle between the dipole moment and the field. At 30 degrees:

\[ \tau = PE\sin 30^{\circ} = \frac{PE}{2} \]

\(\frac{\sqrt{3}}{2}PE\) uses the cosine instead of the sine; the cosine appears in the potential energy, \(U = -PE\cos\theta\), not in the torque. PE is the maximum torque, at 90 degrees.

Answer: C

The potential is inversely proportional to \(r^{2}\).

On the axis of a short dipole of moment \(p\), at distance \(r\) much larger than the dipole length:

\[ V = \frac{1}{4\pi\varepsilon_{0}}\frac{p}{r^{2}} \]

The two charges' potentials nearly cancel, leaving a \(\frac{1}{r^{2}}\) dependence, one power faster than a point charge's \(\frac{1}{r}\).

\(r\) is the point-charge case. \(r^{3}\) belongs to the dipole's electric field, not its potential. \(r^{-1}\) and \(r^{-2}\) would mean the potential grows with distance.