JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 11)
A point dipole $$\overrightarrow p = - {p_0}\widehat x$$
is kept at the origin. The potential and electric field due to this dipole on the
y-axis at a distance d are, respectively: (Take V= 0 at infinity)
$${{\left| {\overrightarrow p } \right|} \over {4\pi { \in _0}{d^2}}},{{ - \overrightarrow p } \over {4\pi { \in _0}{d^3}}}$$
$$0,{{\overrightarrow p } \over {4\pi { \in _0}{d^3}}}$$
$${{\left| {\overrightarrow p } \right|} \over {4\pi { \in _0}{d^2}}},{{\overrightarrow p } \over {4\pi { \in _0}{d^3}}}$$
$$0,{{ - \overrightarrow p } \over {4\pi { \in _0}{d^3}}}$$
Explicació
V = 0
$$E = - {{K\overrightarrow P } \over {{r^3}}}$$
$$ = - {{\overrightarrow p } \over {4\pi {\varepsilon _0}{d^3}}}$$
$$E = - {{K\overrightarrow P } \over {{r^3}}}$$
$$ = - {{\overrightarrow p } \over {4\pi {\varepsilon _0}{d^3}}}$$