Question

Two particles A and B possessing charges of +2.00 × 10⁻⁶ C and of −4.00 × 10⁻⁶ C, respectively, are held fixed at a separation of 20.0 cm. Locate the points (s) on the line AB, where (a) the electric field is zero (b) the electric potential is zero.

Answer 1

Given:
Magnitude of the charges,
$q_1=2.0\times {10}^{-6}\mathrm{C}$ and $q_2=-4.0\times {10}^{-6}\mathrm{C}$
Separation between the charges, r = 0.2 m
(a) Let the electric field be zero at a distance x from A, as shown in the figure.


Then,
$$\begin{gathered} \frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_1}{x^2}+\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_2}{{\left(0.2-\mathrm{x}\right)}^2}=0 \\ \Rightarrow \frac{2}{x^2}+\frac{-4}{{\left(0.2-x\right)}^2}=0 \end{gathered}$$
or x = 48.3 cm from A along BA.

(b) Let the potential be zero at a distance x from A.
$$\begin{gathered} \frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_1}{x}+\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_2}{\left(0.2\pm x\right)}=0 \\ \Rightarrow \frac{2}{x}-\frac{4}{\left(0.2\pm x\right)}=0 \\ \text{Or}\frac{1}{x}=\frac{2}{\left(0.2\pm x\right)} \\ \Rightarrow 0.2\pm x=2x \end{gathered}$$
⇒ x = $\frac{20}{3}\mathrm{cm}$ from A along AB
and x = 20 cm from A along BA.