Test Index

Electrostatics Part 1

Section: Physics

Question : 34 of 100

Marks: +1, -0

Solution:

Electric field one to A and D together will be along –ve x-axis, since, the vertical components get cancelled.

Similarly, due B and C, in the +ve X- axis

Net electric field due to A and D

=2 \times \frac{k q}{D^{2}+4 d^{2}} \times \frac{\hat{D(-i)}}{\sqrt{D^{2}+4 d^{2}}} \ A\ t\ P

Electric field one to B and C at P

=2 \times \frac{k q}{D^{2}+d^{2}} \times \frac{D}{\sqrt{D^{2}+d^{2}}}(\hat{i})

\therefore

Net electric field at

P = \vec{E} = 2 k q D \left( \frac{1}{(D^{2}+d^{2})^{3/2}} - \frac{1}{(D^{2}+d^{2})^{3/2}} \right) \hat{i}

Since, d<<< D,

\vec{E} \approx 2 K q D \left( \frac{1}{D^{3} \left(1+\frac{d^{2}}{D^{2}}\right)^{3/2}} - \frac{1}{D^{2} \left(1+\frac{4 d^{2}}{D^{2}}\right)^{3/2}} \right) \hat{i}

\approx \frac{2 K q D}{D^{3}} \left( 1 - \frac{3}{2} \frac{d^{2}}{D^{2}} - 1 + \frac{3}{2} \times \frac{4 d^{2}}{D^{2}} \right) \hat{i}

\therefore E \propto \frac{1}{D^{4}}

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