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Question

Physics Question on Electric charges and fields

An electric dipole with dipole moment p02(i^+j^)\frac{p_{0}}{\sqrt{2}}\left(\hat{i}+\hat{j}\right) is held fixed at the origin O in the presence of an uniform electric field of magnitude E. 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: (ε0\varepsilon_{0} is permittivity of free space, R >> dipole size)

A

R=(P04πε0E0)1/3R=\left(\frac{P_{0}}{4\pi\varepsilon_{0E_0}}\right)^{^{1/ 3}}

B

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

C

Total electric field at point A is EA=2E0(i^^+j^)\vec{E}_{A}=\sqrt{2}E_{0}\left(\hat{\hat{i}} +\hat{j}\right)

D

Total electric field at point B is EB=0\vec{E}_{B}=0

Answer

Total electric field at point B is EB=0\vec{E}_{B}=0

Explanation

Solution

(1)P=P02(i^+j^)\left(1\right) \overrightarrow{P} = \frac{P_{0}}{\sqrt{2}}\left(\hat{i} +\hat{j}\right)

E.F. at B along tangent should be zero since circle is equipotential.

So, E0=KPR3&EB=0E_{0} = \frac{K\left|\vec{P}\right|}{R^{3}} \& E_{B} = 0

So, R3=KP0E0=(P04π0E0)R^{3} = \frac{KP_{0}}{E_{0}} = \left(\frac{P_{0}}{4\pi\,\in_{0}\,E_{0}}\right)

So R=(P04π0E0)1/3R = \left(\frac{P_{0}}{4\pi \,\in _{0}\,E_{0}}\right)^{1/3}

So, (1) is correct
(2) Because E0E_{0} is uniform & due to dipole E.F. is different at different points, so magnitude of total E.F. will also be different at different points.
So, (2) is incorrect

(3)EA=2KPR3+KPR3=3KPR3P02(i^+j^)\left(3\right) E_{A} = \frac{2KP}{R^{3}} + \frac{KP}{R^{3}} = 3 \frac{KP}{R^{3}} \frac{P_{0}}{\sqrt{2}}\left(\hat{i} + \hat{j}\right)

So,t(3) is wrong

(4)EB=0\left(4\right) E_{B} = 0

so, (4) is correct