InterviewSolution
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InterviewSolution
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Four charge particles each having charge Q=1 C are fixed at the corners of the base(A,B,C, and D) of a square pyramid with slant length `a(AP=BP=PC=a=sqrt(2)m)`, a charge -Q is fixed at point P. A dipole with dipole moment p=1 Cm is placed at the center of the bases and perpendicular to its plane as shown in fig. Force on the dipole due to the charge particles is `(square)/(4 pi epsilon_(0))N`. |
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Answer» Correct Answer - i. `F_("dipole") = (3 sqrt(2) Q q)/(pi in. a^2)` (upward) ii. `U - (Q^2)/(2 sqrt(2) pi epsilon . a) - (p Q)/(2 pi epsilon . a^2)`. a. Charges at A , B, C, and D are placed at an equilateral position of dipole. Hence, the force on each of them due to dipole is `F_1 = (Q p)/(4 pi epsilon_0 (a// sqrt(2))^3)` This force is downward on charges. Henc, force due to these charges on dipole is `4F_1` (upward). Force on dipole due to charge at P is `F_2 = (2 p Q)/(4 pi epsilon_0 (a //sqrt(2))^3)` (upward) Net force on dipole is `F = 4F_1 + F_2 = (3 sqrt (2 Q p))/(pi epsilon_0 a^3)` (upward) b. `PE` of the system is `U = (10 "pairs of charged particles") + (5 "pairs of dipole and charged particles")` As potential energy of the dipole with four charges at `A, B, C, and D` will be zero, `U = 4 [(1)/(4 p epsilon_0) (Q^2)/A] + 2 (1)/(4 pi epsilon_0) (Q^2)/(sqrt(2 a)) - 4(1)/(4 pi epsilon_0) (Q^2)/A` `- (1)/(4 pi epsilon_0) (p Q)/((a// sqrt(2))^2)` `U = (Q^2)/(2 sqrt(2) pi epsilon_0 a)- (p Q)/(2 pi epsilon_0 a^2)`. |
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