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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

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

Grade:12

3 Answers

Yash Sinha
32 Points
13 years ago

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)


Jitendra
13 Points
5 years ago
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
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Potential energy of the system
U = kQ1​Q2/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

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