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From A conducting sphere of radius R a cavity of radius R/2 is cut out as shown in the figure. At the centre of cavity a point charge q is placed then, the potential and field at point P as shown in the figure is

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V=q2πε0d,E=q2πε0d2

V=q4πε0d,E=q2πε0d2

V=q4πε0d,E=q4πε0d2

V=q2πε0d,E=q4πε0d2

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detailed solution

Correct option is C

Charge -q will be induced on the inner surface of the cavity with uniform surface charge density . Also +q charge will be induced on the outer surface of the metallic sphere with uniform surface density . Effect produced by -q and +q at the centre of cavity at point P will cancel outTherefore at point P , E = q4πε0d2 and V = q4πε0d ..

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Similar Questions

Two spherical shells are as shown in figure.

Let r be the distance of a point from their common centre. Then, match the following columns and mark the correct choice from the given codes.

	Column-I		Column-II
i.	Electric field for r < R₁	p.	is constant for q₂ and vary for q₁
ii.	Electric potential for r < R₁	q.	is zero for q₂and vary for q₁
iii.	Electric potential for R₁ < r <R₂	r.	is constant
iv.	Electric field for R₁ < r <R₂	s.	is zero