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Q.

Consider a cube as shown in the figure-I; with uniformly distributed charged in its entire volume. Intensity of electrical field at one of its vertex P is $E_{0}$ . A portion of half the size (half edge length) of the original cube is cut and removed as shown in the figure-II. Then modulus of electric field at the point P in the new structure is $\frac{E_{0}}{n}$ . Find $n = ?$

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Detailed Solution

$E α \frac{Q}{{(L)}^{2}} α \frac{ρ {(L)}^{3}}{{(L)}^{2}}$
$E α L$
$∴ E_{c o m p l e t e} = E$
$E = \frac{E}{8}$ ; $E_{Remaining} = \frac{7 E}{8}$

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Consider a cube as shown in the figure-I; with uniformly distributed charged in its entire volume. Intensity of electrical field at one of its vertex P is E0. A portion of half the size (half edge length) of the original cube is cut and removed as shown in the figure-II. Then modulus of electric field at the point P in the new structure is E0n. Find n=?