Three point charges q, - 2 q and -2 q are placed at the vertices of an equilateral triangle of side a. The work done by some external force to increase their separation to 2a will be

Question

Three point charges q, - 2 q and -2 q are placed at the vertices of an equilateral triangle of side a. The work done by some external force to increase their separation to 2a will be (A) negative (B)  (1/4π ε 0)(2q2/a)  (C)  (1/4π ε 0)(3q2/a)  (D) zero

Tardigrade Admin · Accepted Answer

According to the question work done in increasing the separation from a to 2 a is W=Uf-Ui Here, Ui=(1/4π ε 0)[ (q(-2q)/a)+(q(-2q)/a)+((-2q) (-2q)/a) ] Similarly, Uf is also zero. Hence, W = 0

Q. Three point charges q, - 2 q and -2 q are placed at the vertices of an equilateral triangle of side a. The work done by some external force to increase their separation to 2a will be

negative

$\frac{1}{4 π ε _{0}} \frac{2 q ^{2}}{a}$

$\frac{1}{4 π ε _{0}} \frac{3 q ^{2}}{a}$

zero

According to the question work done in increasing the separation from a to 2 a is $W = U_{f} - U_{i}$
Here, $U_{i} = \frac{1}{4 π ε _{0}} [\frac{q ( - 2 q )}{a} + \frac{q ( - 2 q )}{a} + \frac{( - 2 q ) ( - 2 q )}{a}]$ Similarly, $U_{f}$ is also zero. Hence, W = 0

Q. Three point charges q, - 2 q and -2 q are placed at the vertices of an equilateral triangle of side a. The work done by some external force to increase their separation to 2a will be

negative

4πε0​1​a2q2​

4πε0​1​a3q2​

zero

According to the question work done in increasing the separation from a to 2 a is W=Uf​−Ui​ Here, Ui​=4πε0​1​[aq(−2q)​+aq(−2q)​+a(−2q)(−2q)​] Similarly, Uf​ is also zero. Hence, W = 0

$\frac{1}{4 π ε _{0}} \frac{2 q ^{2}}{a}$

$\frac{1}{4 π ε _{0}} \frac{3 q ^{2}}{a}$

According to the question work done in increasing the separation from a to 2 a is $W = U_{f} - U_{i}$
Here, $U_{i} = \frac{1}{4 π ε _{0}} [\frac{q ( - 2 q )}{a} + \frac{q ( - 2 q )}{a} + \frac{( - 2 q ) ( - 2 q )}{a}]$ Similarly, $U_{f}$ is also zero. Hence, W = 0