Two equal charges q of opposite sign separated by a distance 2 a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining centre to this point makes an angle q with the axis of the dipole, then the potential at the point P is given by (r 2 a)

Question

Two equal charges q of opposite sign separated by a distance 2 a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining centre to this point makes an angle q with the axis of the dipole, then the potential at the point P is given by (r > > 2 a) (A) V=(p cos θ/4 π ε0 r2) (B) V=(p cos θ/4 π ε0 r) (C) V=(p sin θ/4 π ε0 r) (D) V=(p cos θ/4 π ε0 r3)

Tardigrade Admin · Accepted Answer

V=(p cos θ/4 π ε0(r2-a2 cos 2 θ)) =(p cos θ/4 π ε0 r2) when 2 a < < r

$V = \frac{p c o s θ}{4 π ε _{0} r ^{2}}$

$V = \frac{p c o s θ}{4 π ε _{0} r}$

$V = \frac{p s i n θ}{4 π ε _{0} r}$

$V = \frac{p c o s θ}{4 π ε _{0} r ^{3}}$

$V = \frac{p c o s θ}{4 π ε _{0} ( r ^{2} - a ^{2} c o s ^{2} θ )}$
$= \frac{p c o s θ}{4 π ε _{0} r ^{2}}$ when $2 a << r$

V=4πε0​r2pcosθ​

V=4πε0​rpcosθ​

V=4πε0​rpsinθ​

V=4πε0​r3pcosθ​

V=4πε0​(r2−a2cos2θ)pcosθ​ =4πε0​r2pcosθ​ when 2a<<r

$V = \frac{p c o s θ}{4 π ε _{0} r ^{2}}$

$V = \frac{p c o s θ}{4 π ε _{0} r}$

$V = \frac{p s i n θ}{4 π ε _{0} r}$

$V = \frac{p c o s θ}{4 π ε _{0} r ^{3}}$

$V = \frac{p c o s θ}{4 π ε _{0} ( r ^{2} - a ^{2} c o s ^{2} θ )}$
$= \frac{p c o s θ}{4 π ε _{0} r ^{2}}$ when $2 a << r$