CUET Preparation Today
What is the electric potential due to a dipole along with its axial position? |
$\frac{p}{4\pi \epsilon_0 r}$ $\frac{p}{2\pi \epsilon_0 r}$ $\frac{p}{2\pi \epsilon_0 r^2}$ $\frac{p}{4\pi \epsilon_0 r^2}$ |
$\frac{p}{4\pi \epsilon_0 r^2}$ |
$\text{Potential due to a dipole at a point is }V = \frac{p cos\theta}{4\pi\epsilon_0 r^2}$ $\Rightarrow \text{ At axial position }\theta = 0^o \text{ or } 180^o$ $\Rightarrow$ Vaxial = +- \(\frac{p}{4\pi\epsilon_{o}r^{2}}\) |
