Question

Work done in shifting a charge q/2 from a point X to a point Y in the diagram shown in figure is then

• A. $\frac{5(\sqrt{5}-1)K{q}^2}{d\sqrt{5}}$
• B. $\frac{4(\sqrt{5}-1)K{q}^2}{d\sqrt{5}}$
• C. $\frac{6(\sqrt{5}-1)K{q}^2}{d\sqrt{5}}$
• D. $\frac{(\sqrt{5}-1)K{q}^2}{d\sqrt{5}}$

Answer 1

Answer: (B)

$\frac{4(\sqrt{5}-1)K{q}^2}{d\sqrt{5}}$

The potential at point X due to system of system of charges is given by:

$V_x=\frac{k(+q)}{\left(d/2\right)}+\frac{k(+q)}{\left(d/2\right)}+\frac{k(-q)}{\sqrt{d^2+(d/2{)}^2}}+\frac{k(-q)}{\sqrt{d^2+(d/2{)}^2}}$

$\to V_x=\frac{4kq}{d}-\frac{4kq}{d\sqrt{5}}=\frac{4(\sqrt{5}-1)kq}{d\sqrt{5}}$

Similarly potential at point Y is given by:

$V_y=-\frac{4kq}{d}+\frac{4kq}{d\sqrt{5}}=-\frac{4(\sqrt{5}-1)kq}{d\sqrt{5}}$

So work done in moving the charge $q/2$ from X to Y is:

$W=(V_x-V_y)\frac{q}{2}=\frac{4(\sqrt{5}-1)k{q}^2}{d\sqrt{5}}$.