Consider a dipole of moment \(p\) placed in an uniform electrostatic field \(\mathrm{E}_{0} .\) The radius…
- \(\left(k p / \mathrm{E}_{0}\right)^{1 / 3}\)
- \(k p / \mathrm{E}_{0}\)
- \(\left(k p / \mathrm{E}_{0}\right)^{2 / 3}\)
- none of these
Solution
The radius of the equipotential surface is given by \(\frac{k p \sin \theta}{r^{3}}=E_{0} \sin \theta\)\(\ldots(1)\)
\(r=\left(\frac{k p}{E_{0}}\right)^{1 / 3}\)
where \(\mathrm{k}=\left(4 \pi \varepsilon_{0}\right)^{-1}\). Equation (1) follows from the fact that on an equipotential surface the field is at right angles. So, the total tangential component is zero. /
Asked in: JEE Mains - Electrostatics - Chapter Test