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Physics Question on Moving Charges and Magnetism

As per this diagram a point charge +q is placed at the origin O. Work done in taking another point charge -Q from the point A [co-ordinates (0,a)] to another point B [coordinates (a,0)] along the straight path AB is:

a point charge +q is placed at the origin O

A

zero

B

((−qQ4π∈0)(1a2))\bigg((\frac{-qQ}{4\pi\in_0})(\frac{1}{a^2})\bigg) 2\sqrt 2a

C

((qQ4π∈0)(1a2)).a2\bigg((\frac{qQ}{4\pi\in_0})(\frac{1}{a^2})\bigg).a\sqrt 2

D

\bigg((\frac{qQ}{4\pi\in_0})(\frac{1}{a^2})\bigg)$$\sqrt 2a

Answer

zero

Explanation

Solution

Potential energy of two charge system is U =q1q24π∈0r12\frac{q_1q_2}{4\pi \in_0 r_{12}}

Potential energy when charge -Q is at A UA = (−Q)q4π∈0(OA)\frac{(-Q)q}{4\pi \in_0(OA)} = (−Q)q4π∈0a\frac{(-Q)q}{4\pi \in_0a}

Potential energy when charge -Q is at B UB =(−Q)q4π∈0(OB)\frac{(-Q)q}{4\pi \in_0(OB)}= (−Q)q4π∈0a\frac{(-Q)q}{4\pi \in_0a}

Now, work done WAB=-Q(VB-VA)

=UB-UA=(−Q)q4π∈0a\frac{(-Q)q}{4\pi \in_0a} - (−Q)q4π∈0a\frac{(-Q)q}{4\pi \in_0a} = 0

Therefore, the correct option is (A): Zero