Question

electric potential on axial line and equatorial line of an electric dipole derivation​

Answer 1

Answer:

ELECTRIC POTENTIAL OF ELECTRIC DIPOLE AT AXIAL LINE:

-Consider electric dipole at charge '+q' , '-q' and length '2a' such that electric potential is to be determined at point 'P' on Axial line.

Net Electric potential at point 'P' is:

$\mathtt{v_p = v_a + v _b }$

$\mathtt{v_p = \frac{1}{4\pi \in \circ} \times \frac{( - q)}{(r + a)} + \frac{1}{4\pi \in \circ} \times \frac{q}{(r - a)} }$

$\mathtt{v_p = \frac{1}{4\pi \in \circ} \times q \big( \frac{ - r + a + r + a}{(r + a)(r - a)} \big)}$

By Cancelling '-r' from '+r'

$\mathtt{v_p = \frac{q(2a)}{4\pi \in \circ( {r}^{2} - {a}^{2}) } }$

Since, q(2a) = P

$\mathtt{v_p = \frac{p}{4\pi \in \circ( {r}^{2} - {a}^{2}) } }$

For Short Dipole a<<<<<r

$\mathtt{v_p = \frac{p}{4\pi \in \circ{r}^{2} } }$

because a is too much small then r so we can neglect it.

ELECTRIC POTENTIAL OF ELECTRIC DIPOLE AT EQUATORIAL LINE:

$\mathtt{v_p = v_a + v_b}$

$\mathtt{v_p = \frac{1}{4\pi \in \circ} \frac{( - q)}{ \sqrt{ {r}^{2} + {a}^{2} } } + } \frac{1}{4\pi \in \circ} \frac{( q)}{ \sqrt{ {r}^{2} + {a}^{2} } }$

By Cancelling both we get,

$\mathtt{v_p =0}$

therefore electric dipole at equatorial point is zero.