Question

As per the diagram a point charge +q is placed at the origin O. Work done in taking another point charge -Q from the point A [coordinates (0, a)] to another point B [coordinates (1a, 0)] along the straight path AB is

• A. zero
• B. $ \bigg( \frac{ qQ}{ 4 \pi \varepsilon_0} \frac{1}{ a^2} \bigg) \cdot \sqrt 2 a $
• C. $ \bigg( \frac{- qQ}{ 4 \pi \varepsilon_0} \frac{1}{ a^2} \bigg) \cdot \sqrt 2 a $
• D. $ \bigg( \frac{ qQ}{ 4 \pi \varepsilon_0} \frac{1}{ a^2} \bigg) \cdot \frac{a}{ \sqrt 2 } $

Answer 1

Answer: (A)

Work done is equal to zero because the potential of A and B are the same = $ \frac{ 1}{ 4 \pi \varepsilon_0} \frac{q}{a}$

No work is done if a particle does not change its potential energy.
i.e. initial potential energy = final potential energy.