Coulombs Law and Electric Field Questions and Answers

Answer: B

The force on the dipole is in the +y direction.

Place the dipole at (d, 0, 0). The charge q sits on the dipole's perpendicular bisector (equatorial line), where the dipole's field is antiparallel to the dipole moment:

\[ \vec{E}_{dip} = -\frac{p}{4\pi\epsilon_{0}d^{3}}\,\hat{y} \]

So the dipole pushes the positive charge in −y, and by Newton's third law the charge pushes the dipole in +y with force \(qp/(4\pi\epsilon_{0}d^{3})\). The same follows from \(\vec{F} = (\vec{p} \cdot \vec{\nabla})\vec{E}\): moving in y off the x axis, the charge's field gains a +y component.

The force is not zero, because the field is non-uniform. It has no x component, since the dipole's two charges sit symmetrically about the x axis at equal distances from q.

Answer: C

The ratio \(E_{P}/E_{Q}\) is 4 : 1.

An infinite line charge produces a field perpendicular to the line with magnitude \(E = \frac{\lambda}{2\pi\epsilon_{0} d}\), so the field is inversely proportional to the perpendicular distance d from the line.

P is at distance a from AB. In the figure, PQ = 5a and the horizontal leg is 4a, so the vertical rise from P to Q is

\[ \sqrt{(5a)^{2} - (4a)^{2}} = 3a \]

Q is therefore at \(a + 3a = 4a\) from the line, and

\[ \frac{E_{P}}{E_{Q}} = \frac{4a}{a} = 4 \]

1 : 4 inverts the ratio, and 3 : 1 or 1 : 3 use the 3a rise instead of Q's full distance 4a from the line.

Answer: D

The dipole moment is 4q.

The net charge is \(q + q - 2q = 0\), so the dipole moment does not depend on the origin and is \(\bar{p} = \sum q_{i}\bar{r}_{i}\):

\[ \begin{aligned} \bar{p} &= q(0, -1, 1) + q(0, 1, 1) \\ &\quad - 2q(0, 0, -1) \\ &= (0,\ 0,\ 2q + 2q) = (0, 0, 4q) \end{aligned} \]

The y-components of the two positive charges cancel, and their z-components (q each) add to the \(+2q\) from the negative charge at \(z = -1\). Answers of 2q or zero come from forgetting the sign of the \(-2q\) charge's position or counting only one part.

Answer: C

The field of an ideal dipole falls off as \(r^{-3}\).

The potential of a dipole p at large distance is

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

and the field, its negative gradient, gains one more power of r in the denominator:

\[ \begin{aligned} E &= \frac{p}{4\pi\varepsilon_{0}r^{3}} \\ &\quad\times\sqrt{3\cos^{2}\theta + 1} \end{aligned} \]

The fields of the two equal and opposite charges nearly cancel, which is why the field drops faster than the \(r^{-2}\) field of a single charge. \(r^{-2}\) is the dipole potential's behaviour, not the field's, and \(r^{-4}\) is the field of a quadrupole.

Answer: C

The force is greatest when the charges are Q/2 and Q/2.

With parts q and Q - q at a fixed distance d, the Coulomb force is

\[ F = \frac{q(Q - q)}{4\pi\varepsilon_{0}d^{2}} \]

Setting \(dF/dq = 0\) gives \(Q - 2q = 0\), so q = Q/2, and the second derivative is negative, so this is a maximum. For example, Q/2 and Q/2 give a product \(Q^{2}/4\), while Q/4 and 3Q/4 give only \(3Q^{2}/16\) and Q/3 and 2Q/3 give \(2Q^{2}/9\). Since the product changes with the split, 'any q1 and q2' is wrong too.

Answer: D

The field is zero for \(|x| \lt a\), \(\frac{\sigma}{\varepsilon_{0}}\hat{x}\) for \(x \gt a\) and \(-\frac{\sigma}{\varepsilon_{0}}\hat{x}\) for \(x \lt -a\).

Each infinite sheet of charge density \(\sigma\) produces a uniform field \(\sigma/2\varepsilon_{0}\) pointing away from it on both sides. Superpose the two:

• Between the plates, the sheet at \(-a\) pushes along \(+\hat{x}\) and the sheet at \(+a\) along \(-\hat{x}\); they cancel, so \(\vec{E} = 0\).
• For \(x \gt a\), both push along \(+\hat{x}\): \(\vec{E} = \sigma/\varepsilon_{0}\,\hat{x}\).
• For \(x \lt -a\), both push along \(-\hat{x}\): \(\vec{E} = -\sigma/\varepsilon_{0}\,\hat{x}\).

The options with zero field outside and a non-zero field inside describe a pair of oppositely charged plates, a capacitor, not two positive sheets.

Answer: A

The force is \( F = Q_1 Q_2/(4\pi\epsilon_0 R^2) \), directed along the line joining the charges and repulsive for like signs. The inverse square dependence is what makes Gauss's law take so simple a form in three dimensions.

Answer: B

Correct answer: \( 9 \times 10^{9} \)

With \( \epsilon_0 = 8.85\times10^{-12} \) F/m, \[ \frac{1}{4\pi\epsilon_0} = 8.99\times10^{9} \approx 9\times10^{9}\ \mathrm{N\,m^2\,C^{-2}}. \] The 8.85 offered is the permittivity itself rather than the constant built from it, \( 3\times10^{8} \) is the speed of light, and \( 4\pi\times10^{-7} \) is the permeability of free space.

Answer: C

It appears in the capacitance of a parallel plate capacitor as \( \epsilon_0 A/d \), where the area over the separation is a length, so the constant itself must be a capacitance per unit length. Its value is \( 8.854 \times 10^{-12} \) F/m.

Answer: D

Coulomb's law gives a positive product for like charges and the vector form then points each charge away from the other. The force acts along the line of centres, which is what makes it a central force and conserves angular momentum about either charge.