JEE Advanced · Physics · 17. Electrostatics
Positive and negative point charges of equal magnitude are kept at \(\left(0,0, \frac{a}{2}\right)\) and \(\left(0,0, \frac{-a}{2}\right)\), respectively. The work done by the electric field when another positive point charge is moved from \((-a, 0,0)\) to \((0, a, 0)\) is
- A positive
- B negative
- C zero
- D depends on the path connecting the initial and final positions
Answer & Solution
Correct Answer
(C) zero
Step-by-step Solution
Detailed explanation
\(A \equiv(-a, 0,0) B \equiv(0, a, 0)\)
Point charge is moved from \(A\) to \(B\)
\(
\begin{aligned}
V_A & =V_B=0 \\
W & =0
\end{aligned}
\)
\(
\therefore \quad W=0
\)
or the correct option is (c).

Point charge is moved from \(A\) to \(B\)
\(
\begin{aligned}
V_A & =V_B=0 \\
W & =0
\end{aligned}
\)
\(
\therefore \quad W=0
\)
or the correct option is (c).

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