An electric dipole is as shown in figure The electric potential at point $\mathrm{P}$ due to the dipole is…

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]
  1. $\frac{a q}{2 \pi \epsilon_0\left(x^2+a^2\right)}$
  2. $\frac{2 a q}{2 \pi \epsilon_0\left(x^2-a^2\right)}$
  3. $\frac{2 a q}{2 \pi \epsilon_0\left(x^2+a^2\right)}$
  4. $\frac{\mathrm{aq}}{2 \pi \epsilon_0\left(\mathrm{x}^2-\mathrm{a}^2\right)}$

Solution

$\begin{aligned} & \mathrm{V}_{-\mathrm{q}}=\frac{1}{4 \pi \varepsilon_0} \frac{(-\mathrm{q})}{(\mathrm{x}+\mathrm{a})} \\ & \mathrm{V}_{+\mathrm{q}}=\frac{1}{4 \pi \varepsilon_0} \frac{(+\mathrm{q})}{(\mathrm{x}-\mathrm{a})} \\ & \therefore \mathrm{V}_{\mathrm{d}}=\frac{1}{4 \pi \varepsilon_0} \mathrm{q} \frac{((\mathrm{x}+\mathrm{a})-(\mathrm{x}-\mathrm{a}))}{\left(\mathrm{x}^2-\mathrm{a}^2\right)}=\frac{1}{2 \pi \varepsilon_0} \frac{(\mathrm{aq})}{\left(\mathrm{x}^2-\mathrm{a}^2\right)}\end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 2)

Practice more Electrostatics questions on Aicharya