Question

Points A and B are 0.10 m apart. A point charge of +3.0 x 10-9 C is placed at A and a

point charge of -1.0 x 10-9 C is placed at B.

I. X is the point on the straight line through A and B, between A and B, where the

electric potential is zero. Calculate the distance AX.

II. Show on a diagram the approximate position of a point, Y, on the straight line

through A and B where the electric field strength is zero.( no calculation is

expected)

Answer 1

Explanations & Calculations

• The Potential of a point due to a point charge is given by $\small V=\large\frac{kq}{r}$

1)

• Therefore, the potential of point X due to the given two charges is

$\qquad\qquad \begin{aligned} \small V_x&=\small \Sigma V_i = V_{q_1}+V_{q_2}\\ &=\small \frac{kq_1}{AX}+\frac{kq_2}{(0.1-AX)}\\ &=\small k\bigg[\frac{3\times10^{-9}C}{AX}+\frac{(-1.0\times10^{-9}C)}{(0.1-AX)}\bigg] \end{aligned}$

• Since $\small V_x$ is zero,

$\qquad\qquad \begin{aligned} \small \frac{3\times10^{-9}}{AX}&=\small \frac{1.0\times10^{-9}}{(0.1-AX)}\\ \small 0.3-3AX&=\small AX\\ \small AX&=\small \frac{0.3}{4}\Big[=\frac{3}{4}(0.1m)\Big]\\ &=\small \bold{0.075m} \end{aligned}$

• The point of zero potential is located more closely to the weaker charge.

2)

• Field strength is the force on a unit charge.
• And according to the equation $\small \vec{E}=\large \frac{k|q|}{r^2}$, the force\field strength is stronger near the source charge and fades away with the distance.
• Since the given two charges are of the opposite kind, force on a unit charge is also in both directions.
• Therefore, a good counter force\field to the one due to the larger charge ($\small 3\,nC$), is possible by the weaker charge near itself.
• As shown in the figure, regions A and C have the possibility of bearing a zero-field point (depends on the magnitudes of the two charges); but for this case, it is only region C.

• At the point (Y- described in the question) of zero fields, magnitudes of the positive portion & the negative portion are equal.