MHT CET · Physics · Electrostatics
Three charges each of value \(+q\) are placed at the corners of an isosceles triangle \(\mathrm{ABC}\) of sides \(\mathrm{AB}\) and \(\mathrm{AC}\) each equal to \(2 \mathrm{a}\). The mid points of \(A B\) and \(A C\) are \(D\) and \(E\) respectively. The work done in taking a charge \(\mathrm{Q}\) from \(\mathrm{D}\) to \(\mathrm{E}\) is \(\left(\varepsilon_0=\right.\) permittivity of free space)
- A Zero
- B \(\frac{3 \mathrm{qQ}}{4 \pi \varepsilon_0 \mathrm{a}}\)
- C \(\frac{\mathrm{qQ}}{8 \pi \varepsilon_0 \mathrm{a}}\)
- D \(\frac{3 \mathrm{qQ}}{8 \pi \varepsilon_0 \mathrm{a}}\)
Answer & Solution
Correct Answer
(A) Zero
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \text { Given } A B=A C=2 a \\
& V_D=V_E \quad(\because D \text { and } E \text { are mid-points })
\end{aligned}
\)
\(\therefore \quad\) The work done in taking a charge \(\mathrm{q}\) from \(\mathrm{D}\) to \(\mathrm{E}\) is
\(
\begin{array}{rlrl}
& W=q \Delta V=q\left(V_D-V_E\right) \\
\therefore& W=0 \text { (Equipotential surfaces) }
\end{array}
\)
\begin{aligned}
& \text { Given } A B=A C=2 a \\
& V_D=V_E \quad(\because D \text { and } E \text { are mid-points })
\end{aligned}
\)
\(\therefore \quad\) The work done in taking a charge \(\mathrm{q}\) from \(\mathrm{D}\) to \(\mathrm{E}\) is
\(
\begin{array}{rlrl}
& W=q \Delta V=q\left(V_D-V_E\right) \\
\therefore& W=0 \text { (Equipotential surfaces) }
\end{array}
\)
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