Question

compute electric potential at a distance 0.1 metre from a point charge of 4×10^⁹ coulomb​

Answer 1

Given ,

Charge (q) = 4 × 10⁹ C

Distance (r) = 0.1 m

We know that , the electric potential due to point charge at point r is given by

$\boxed{ \tt{v = \frac{kq}{r}} }$

Thus ,

$\sf \mapsto v = \frac{9 \times {(10)}^{9} \times 4 \times {(10)}^{9} }{0.1}$

$\sf \mapsto v = 3.6 \times {(10)}^{20} \: \: volt$

∴ The electric potential is 3.6 × 10^(20) v