Question

Electric potential at a point ‘P’ due to a point charge of 5 × 10⁻⁹ C is 50 V. The distance of ‘P’ from the point charge is ______.

(Assume, `1/(4piε_0)` = 9 × 10⁺⁹ Nm² C⁻²)

Options

• 3 cm

• 9 cm

• 90 cm

• 0.9 cm

Answer 1

Electric potential at a point ‘P’ due to a point charge of 5 × 10⁻⁹ C is 50 V. The distance of ‘P’ from the point charge is 90 cm.

Explanation:

Given: Charge (q) = 5 × 10⁻⁹ C

Electric Potential (V) = 50 V
Electrostatic constant k = `1/(4piε_0)` = 9 × 10⁺⁹ Nm² C⁻² 

Formula: Electric Potential due to a point charge is:

V = `(kq)/r`

r = `(kq)/V`

= `((9 xx 10^9) xx (5 xx 10^-9))/50`

= `45/50`

= 0.9 m

= 0.9 × 100

= 90 cm