Question

The electric potential at a distance of 20 cm from a point charge of 10 microcoulomb

Answer 1

Answer:

$\boxed{\mathfrak{Electric \ potential = 2.25 \ MN/C}}$

Explanation:

Charge (q) = 10 μC = $\sf 10 \times 10^{-6}$ C

Distance (r) = 20 cm = 0.2 m

Electric potential (E) due to a point charge is given as:

$\boxed{ \bold{E = \dfrac{kq}{r^2}}}$

k → Coulomb's constant ($\rm 9 \times 10^9$ Nm²/C²)

By substituting values in the formula we get:

$\rm \implies E = \dfrac{9 \times {10}^{9} \times 10 \times {10}^{ - 6} }{(0.2)^2} \\ \\ \rm \implies E = \frac{9 \times {10}^{9 - 5} }{4 \times {10}^{ - 2} } \\ \\ \rm \implies E = 2.25 \times {10}^{4 + 2} \\ \\ \rm \implies E = 2.25 \times {10}^{6} \: N/C\\ \\ \rm \implies E = 2.25 \: MN/C$