MHT CET · Physics · Electrostatics
Four electric charges \(+q,+q,-q\) and \(-q\) are placed in order at the corners of a square of side \(2 r\). The electric potential at a point midway between the two negative charges is
- A \(\frac{1}{4 \pi \varepsilon_0} \frac{2 q}{r}[1-\sqrt{5}]\)
- B \(\frac{1}{4 \pi \varepsilon_0} \frac{2 q}{r}\left[\frac{1}{\sqrt{5}}+1\right]\)
- C \(\frac{1}{4 \pi \varepsilon_0} \frac{2 q}{r}\left[\frac{1}{\sqrt{5}}-1\right]\)
- D Zero
Answer & Solution
Correct Answer
(C) \(\frac{1}{4 \pi \varepsilon_0} \frac{2 q}{r}\left[\frac{1}{\sqrt{5}}-1\right]\)
Step-by-step Solution
Detailed explanation
Consider the following figure:

\(D P=P C=r\) and \(A P=B P=\sqrt{A D^2+D P^2}=\sqrt{r^2+(2 r)^2}=\sqrt{5} r\)
Potential at \(P\) is \(V_P=\frac{1}{4 \pi \varepsilon_0}\left[\frac{q}{A P}+\frac{-q}{C P}+\frac{-q}{D P}+\frac{q}{B P}\right]\)
\(\Rightarrow V_A=\frac{1}{4 \pi \varepsilon_0} V_A\left[\frac{-q}{r}+\frac{+q}{r \sqrt{5}}+\frac{+q}{r \sqrt{5 L}}+\frac{-q}{r}\right]\) \(=\frac{1}{4 \pi \varepsilon_0} \frac{2 q}{r}\left(\frac{1}{\sqrt{5}}-1\right)\)

\(D P=P C=r\) and \(A P=B P=\sqrt{A D^2+D P^2}=\sqrt{r^2+(2 r)^2}=\sqrt{5} r\)
Potential at \(P\) is \(V_P=\frac{1}{4 \pi \varepsilon_0}\left[\frac{q}{A P}+\frac{-q}{C P}+\frac{-q}{D P}+\frac{q}{B P}\right]\)
\(\Rightarrow V_A=\frac{1}{4 \pi \varepsilon_0} V_A\left[\frac{-q}{r}+\frac{+q}{r \sqrt{5}}+\frac{+q}{r \sqrt{5 L}}+\frac{-q}{r}\right]\) \(=\frac{1}{4 \pi \varepsilon_0} \frac{2 q}{r}\left(\frac{1}{\sqrt{5}}-1\right)\)
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