Find the electrical potential at a point if 0.002 J work is to be done to bring a 40 μC charge from infinity to that point.

Charge, q = 40 μC = $40\times {10}^{-6}\mathrm{C}$. 

Work done, W = 0.002 J
So, 
Electrical potential, 

$\mathrm{V}=\frac{\mathrm{work} \mathrm{done}}{\mathrm{charge}}$
$\Rightarrow \mathrm{V}=\frac{0.002}{40\mathrm{ }\times \mathrm{ }{10}^{-6}\mathrm{ }}$ 

Multiplying numerator and denominator with 10⁹

$\Rightarrow \mathrm{V}=\frac{2\mathrm{ }\times \mathrm{ }{10}^6}{40\mathrm{ }\times \mathrm{ }1000\mathrm{ }}$ 

$\Rightarrow \mathrm{V}=\frac{200}{4\mathrm{ }}$ 

$\Rightarrow \mathrm{V}=50 \mathrm{volt}$ 

So, the electrical potential at that point is $50\mathit{ V}$.