Question 831: Electrostatics & Capacitance

Question

An electric dipole with dipole moment ${{{p_0}} \over {\sqrt 2 }}(\widehat i + \widehat j)$ is held fixed at the origin O in the presence of a uniform electric field of magnitude E
0
.
If the potential is constant on a circle of radius R centered at the origin as shown in figure, then the correct statement(s) is/are, ($ \in $
0
is the permittivity of the free space, R >> dipole size)
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l71eqn0o/2bdc0ca3-130a-459f-bf90-91889254b6be/8ee2e170-2041-11ed-9b51-3bcb6071cf1a/file-1l71eqn0p.png?format=png)

Accepted Answer

R >> Dipole size.
Circle is equipotential.
So, E
net
should be perpendicular to surface hence,
${{k{p_0}} \over {{r^3}}} = {E_0}$
$ \Rightarrow $ $R = {\left( {{{k{p_0}} \over {{E_0}}}} \right)^{1/3}}$
At point B, net electric field will be zero.
E
B
= 0
${({E_A})_{net}} = {{2k{p_0}} \over {{R^3}}} + {E_0} = 3{E_0}$
Electric field at point A, ${E_A} = {3 \over {\sqrt 2 }}{E_0}[\widehat i + \widehat j]$
${({E_B})_{net}}$ = 0
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l71erwcm/5a813f6b-3be4-4855-98ff-39d845f3cd59/b1e79260-2041-11ed-bbab-51526ca2050d/file-1l71erwcn.png?format=png)

Answer

The magnitude of total electric field on any two points of the circle will be same.

Answer

Total electric field at point B is ${\overrightarrow E _B}$ = 0

Answer

$R = {\left( {{{{p_0}} \over {4\pi { \in _0}{E_0}}}} \right)^{1/3}}$

Answer

Total electric field at point A is
${\overrightarrow E _A} = \sqrt 2 {E_0}(\widehat i + \widehat j)$

Electrostatics & Capacitance

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