MHT CET · Physics · Electrostatics
An electric dipole is as shown in figure The electric potential at point \(\mathrm{P}\) due to the dipole is \(\left[\epsilon_0=\right.\) permittivity of free space]

- A \(\frac{a q}{2 \pi \epsilon_0\left(x^2+a^2\right)}\)
- B \(\frac{2 a q}{2 \pi \epsilon_0\left(x^2-a^2\right)}\)
- C \(\frac{2 a q}{2 \pi \epsilon_0\left(x^2+a^2\right)}\)
- D \(\frac{\mathrm{aq}}{2 \pi \epsilon_0\left(\mathrm{x}^2-\mathrm{a}^2\right)}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{aq}}{2 \pi \epsilon_0\left(\mathrm{x}^2-\mathrm{a}^2\right)}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{V}_{-\mathrm{q}}=\frac{1}{4 \pi \varepsilon_0} \frac{(-\mathrm{q})}{(\mathrm{x}+\mathrm{a})}\)
\(\mathrm{V}_{+\mathrm{q}}=\frac{1}{4 \pi \varepsilon_0} \frac{(+\mathrm{q})}{(\mathrm{x}-\mathrm{a})}\)
\(\therefore \mathrm{V}_{\mathrm{d}}=\frac{1}{4 \pi \varepsilon_0} \mathrm{q} \frac{((\mathrm{x}+\mathrm{a})-(\mathrm{x}-\mathrm{a}))}{\left(\mathrm{x}^2-\mathrm{a}^2\right)}\) \(=\frac{1}{2 \pi \varepsilon_0} \frac{(\mathrm{aq})}{\left(\mathrm{x}^2-\mathrm{a}^2\right)}\)
\(\mathrm{V}_{+\mathrm{q}}=\frac{1}{4 \pi \varepsilon_0} \frac{(+\mathrm{q})}{(\mathrm{x}-\mathrm{a})}\)
\(\therefore \mathrm{V}_{\mathrm{d}}=\frac{1}{4 \pi \varepsilon_0} \mathrm{q} \frac{((\mathrm{x}+\mathrm{a})-(\mathrm{x}-\mathrm{a}))}{\left(\mathrm{x}^2-\mathrm{a}^2\right)}\) \(=\frac{1}{2 \pi \varepsilon_0} \frac{(\mathrm{aq})}{\left(\mathrm{x}^2-\mathrm{a}^2\right)}\)
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