From an insulating solid hemisphere (radius R) carrying a charge Q, a part of it is removed in the possibilities shown in List - I. List - II gives the magnitude of the electric field at the centre (O

Question

From an insulating solid hemisphere (radius R) carrying a charge Q, a part of it is removed in the possibilities shown in List - I. List - II gives the magnitude of the electric field at the centre (O) of the original solid hemisphere for the four combinations shown. Match List - I with List - II. (given $K = \frac{1}{4 π ε _{0}}$ )

List - I	List - II
(P)	(1) $0$
(Q)	(2) $\frac{3 K Q}{4 R ^{2}}$
(R)	(3) $\frac{11 K Q}{18 R ^{2}}$
(S)	(4) $\frac{K Q}{2 R ^{2}}$
	(5) $\frac{25 K Q}{18 R ^{2}}$

TestHub · Accepted Answer

For a uniform solid hemisphere the field at its centre is $E_{0} = ρ K π R = \frac{3 K Q}{2 R ^{2}}$ along the axis; for the same density it scales with the radius. A full sphere of radius r contributes $\frac{4}{3} K π ρ r$ at its own surface.

(P) Removed full sphere (R/2) centred R/2 from O on the axis; O lies on its surface: $E = E_{0} - \frac{2}{3} ρ K π R = \frac{E _{0}}{3} = \frac{K Q}{2 R ^{2}}$ - 4.

(Q) The removed sphere is placed touching the pole, its centre $\frac{3 R}{2}$ from O, adding $\frac{K ρ \cdot \frac{4 π}{3} ( R /2 ) ^{3}}{( 3 R /2 ) ^{2}} = \frac{2}{27} ρ K π R$ : $E = E_{0} (1 - \frac{2}{3} + \frac{2}{27}) = \frac{11}{27} E_{0} = \frac{11 K Q}{18 R ^{2}}$ - 3.

(R) Removed hemisphere (R/2) centred at O: $E = E_{0} - \frac{ρ K π R}{2} = \frac{E _{0}}{2} = \frac{3 K Q}{4 R ^{2}}$ - 2.

(S) The removed hemisphere is placed symmetrically on the other side of the base plane and cancels the removed part's field: $E = E_{0} - \frac{E _{0}}{2} - \frac{E _{0}}{2} = 0$ - 1.

Hence P-4, Q-3, R-2, S-1 - (A).

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