ExamBro
ExamBro
MHT CET · Physics · Electrostatics

In the electric field due to a charge ' \(Q\) ', a charge ' \(q\) ' moves from point A to B. The work done is ( \(\varepsilon_0=\) permittivity of vacuum)

  1. A \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathbf{r}^2}\)
  2. B \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}} \times \frac{\pi}{6}\)
  3. C \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}}\)
  4. D Zero
Verified Solution

Answer & Solution

Correct Answer

(D) Zero

Step-by-step Solution

Detailed explanation

The points A and B are equipotential surfaces as both are at the same distance from the charge Q. Therefore, the work done is zero.
Same subject
Explore more questions on app