MHT CET · Physics · Work Power Energy
In an electric field due to charge \(Q\), a charge \(q\) moves from point \(A\) to \(B\) as shown in the figure. The work done is ( \(\varepsilon_0=\) permittivity of free space)
- A \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}^2}\)
- B \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}^2} \frac{\pi}{6}\)
- C \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}}\)
- D zero
Answer & Solution
Correct Answer
(D) zero
Step-by-step Solution
Detailed explanation
Electric field is perpendicular to the displacement.
\(\overrightarrow{\mathrm{F}}=\mathrm{q} \overrightarrow{\mathrm{E}}\)
\(\therefore \vec{F}\) is also perpendicular to the curved surface
\(\mathrm{W} =\int \mathrm{dW} \)
\( =\int \overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{ds}}=\mathrm{F} \cdot \mathrm{ds} \cos 90^{\circ} \)
\( \therefore \quad \mathrm{W} =0\)
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