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Question
Two particles A and B, each carrying a charge Q, are held fixed with a separation dbetween them. A particle C of mass m and charge q is kept at the middle point of the line AB. If it is displaced through a distance x perpendicular to AB, what would be the electric force experienced by it?
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Solution
The charge q is displaced by a distance x on the perpendicular bisector of AB.
As shown in the figure, the horizontal component of the force is balanced.
\[\sin\theta = \frac{x}{\sqrt{\left( \frac{d}{2} \right)^2 + x^2}}\]
Total vertical component of the force,
\[F' = 2F\sin\theta\]
\[F' = 2 \times \frac{1}{4\pi \epsilon_0} \times \frac{qQ}{\left( \frac{d}{2} \right)^2 + x^2} \times \frac{x}{\sqrt{\left( \frac{d}{2} \right)^2 + x^2}}\]
\[ \Rightarrow F' = \frac{1}{2\pi \epsilon_0} \times \frac{qQx}{\left[ \left( \frac{d}{2} \right)^2 + x^2 \right]^{3/2}}\]
This is the net electric force experienced by the charge q.
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