Consider a cube as shown in the figure1 : with uniformly distributed charge in its entire volume. Intensity of electrical field and potential at one of its vertex P are $E_{0}$ and $V_{0}$ respectively. A portion of half the size (half edge length) of the original cube is cut and removed as shown in the figure II. Find modulus of electric field and potential at the point P in the new structure (diagrams are not to scale )

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$E_{p} = \frac{E_{0}}{2}$

$V_{p} = \frac{3 V_{0}}{4}$

$E_{p} = \frac{3 E_{0}}{4}$

$V_{p} = \frac{V_{0}}{4}$

answer is A, B.

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Detailed Solution

Let $E = \frac{k Q}{l^{2}}$ where k is
Some constant
(E is proportional to Q and inversely proportional to dimension squared)
For the removed cube
$E_{r e m o v e d} = \frac{k \frac{Q}{8}}{{(\frac{l}{2})}^{2}} = \frac{k Q}{2 l^{2}}$
$E_{n e t} = E - \frac{E}{2} = \frac{E}{2}$
Similarly $l e t v = \frac{k Q}{l}$
$V_{r e m o v e d} = \frac{k \frac{Q}{8}}{\frac{l}{2}} = \frac{k Q}{4 l}$
$V_{n e t} = V - \frac{V}{4} = \frac{3 V}{4}$