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Physics Question on electrostatic potential and capacitance

Electric field on the axis of a small electric dipole at a distance rr is E→1\overrightarrow{ E }_{1} and E→2\overrightarrow{ E }_{2} at a distance 2r2 r on a line of perpendicular bisector. Then

A

E→2=−E→18\overrightarrow{ E }_{2}=-\frac{\overrightarrow{ E }_{1}}{8}

B

E⃗2=−E⃗116\vec{E}_{2}=-\frac{\vec{E}_{1}}{16}

C

E→1=−E→28\overrightarrow{ E }_{1}=-\frac{\overrightarrow{ E }_{2}}{8}

D

E→1=E→216\overrightarrow{ E }_{1}=\frac{\overrightarrow{ E }_{2}}{16}

Answer

E⃗2=−E⃗116\vec{E}_{2}=-\frac{\vec{E}_{1}}{16}

Explanation

Solution

E→2=−E→116\quad \overrightarrow{ E }_{2}=-\frac{\overrightarrow{ E }_{1}}{16} For axis E1=kx2pr2E _{1}=\frac{ k x 2 p }{ r ^{2}} For bisector E2=−kp(2r)3=kp8r3…E _{2}=-\frac{ k p }{(2 r )^{3}}=\frac{ k p }{ 8 r ^{3}} \ldots (2) E2=−18×kpr3E _{2}=-\frac{1}{8} \times \frac{ k p }{ r ^{3}} =−18×22=-\frac{1}{8} \times \frac{2}{2} =−18×2×E1=-\frac{1}{8 \times 2} \times E _{1} E2=−E116E _{2}=-\frac{ E _{1}}{ 1 6 }