Question

Two charges 5 × 10⁻⁸ C and −3 × 10⁻⁸ C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Answer 1

Given, q₁ = 5 × 10^-8 C, q₂ = −3 × 10^-8 C

And r = 16 cm = 16 × 10^-2 m = 0.16 m

Let the electric potential at point P on the line joining the two charges be zero. If the distance of P from q₁ is x, then the distance of P from q₂ will be (r - x).

Electric potential at P-

Formula: V = `1/(4piε_0) q/r`

V₁ due to q₂ = `(9 xx 10^9 xx 5 xx 10^-8)/x`

and V₂ due to q₂ = `(9 xx 10^9 xx (-3 xx 10^-8))/((r - x))`

Therefore total potential at P, V = V₁ + V₂ = 0

Or `(9 xx 10^9 xx 5 xx 10^-8)/x + (9 xx 10^9 xx (-3 xx 10^-8))/(r - x) = 0`

Or `5/x - 3/((0.16 - x)) = 0`

Or `5/x = 3/(0.16 - x)`

Or 5 (0.16 - x) = 3x

Or 0.8 - 5x = 3x

Or 8x = 0.8 m

∴ `x = 0.8/8`

= 0.1 m

The electric potential will also be zero at a point P external to the line joining the two charges,

let, BP₁ = x 

P₁ at q₁ the potential V'₁ due to = `1/(4piε_0) ⋅ (5 xx 10^-8)/(0.16 + x)`

and potential V'₂ due to q₂ = `1/(4piε_0) · (-3 xx 10^-8)/x`

P₁ Resultant potential V' = V'₁ + V'₂ = 0

Or `1/(4piε_0) (5 xx 10^-8)/(0.16 + x) + 1/(4piε_0) (-3 xx 10^-8)/x = 0`

Or `1/(4piε_0) (5 xx 10^-8)/(0.16 + x) = 1/(4piε_0) (-3 xx 10^-8)/x`

Or `5/(0.16 + x) = 3/x`

Or 5x = 0.48 + 3x

Or 2x = 0.48 m

Or x = 0.24 m

= 24 m