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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.
Solution
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`
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