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, 

\[q_1  = 2 . 0 \times  {10}^{- 6}   C\]  and

\[q_2  =  - 4 . 0 \times  {10}^{- 6}   C\]

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,

\[\frac{1}{4\pi \epsilon_0}\frac{q_1}{x^2} + \frac{1}{4\pi \epsilon_0}\frac{q_2}{\left( 0 . 2 - x \right)^2} = 0\] 

\[ \Rightarrow \frac{2}{x^2} + \frac{- 4}{\left( 0 . 2 + x \right)^2} = 0\]

or x = 48.3 cm from A along BA.

(b) Let the potential be zero at a distance x from A.

\[\frac{1}{4\pi \epsilon_0}\frac{q_1}{x} + \frac{1}{4\pi \epsilon_0}\frac{q_2}{\left( 0 . 2 - x \right)} = 0\] 

\[ \Rightarrow   \frac{2}{x} - \frac{-4}{\left( 0 . 2- x \right)} = 0\] 

\[\text{ Or }  \frac{1}{x} = \frac{2}{\left( 0 . 2 - x \right)}\] 

\[ \Rightarrow 0 . 2 - x = 2x\] 

= 2x = 2.20 - x

= 3x = 0.20

= `x = 0.20/3`