WBJEE · Physics · Electrostatics
Two charges, each equal to \(-q\) are kept at \((-a, 0)\) and \((a, 0)\). A charge \(q\) is placed at the origin. If \(q\) is given a small displacement along y direction, the force acting on q is proportional to
- A \(y\)
- B -y
- C \(\frac{1}{y}\)
- D \(-\frac{1}{y}\)
Answer & Solution
Correct Answer
(B) -y
Step-by-step Solution
Detailed explanation
\(\vec{F}=+q\) (Electric field due to the two negative charges \(=\vec{E}\) ) \[ \vec{F}=+q\left[\frac{k(-2 q) \vec{y}}{a^3}\right] \Rightarrow \vec{F}=-\frac{2 k q^2 y}{a^3} \Rightarrow \vec{F} \propto-\vec{y} \]
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