Electrostatics> Q 17 screening JEE 2000...

Three charges Q, +q and –q are p;aced at the vertices of a right angle triangle (isosceles triangle) as shown. The net electrostatic energy of the configuration is zero if Q is equal to

(A) (-q)/(1+√2)
(B) (-2q)/(2+√2)
(C) -2q
(D) +q

sahil aggarwal , 14 Years ago

Grade 12

3 Answers

Yash Sinha

net electrostatic energy of the configuration = k[ ( q/a) + (Q/a) + (q/a√2) ] = 0

or, q(1+(1/√2))=-Q

Q=

(-q)/(1+√2)

Last Activity: 14 Years ago

Jitendra

Three charge Q , +q and +q are placed at the vertices of a right angle isosceles triangle as shown in figure . The net electrostatic energy of the configuration is zero.Q is equal to

Last Activity: 7 Years ago

Rishi Sharma

Dear Student,
Please find below the solution to your problem.

Potential energy of the system
U = kQ1Q2/r
Where, k = 1/4πε0
Potential energy of the configuration
U = kQq/a + kq^2/a + kqQ/a(2)^0.5
Q = (2)^0.5 q/ ((2)^0.5 + 1)
Hence, Qis equal to (2)^0.5 q/ ((2)^0.5 + 1)

Thanks and Regards

Last Activity: 5 Years ago