Question

ASK YOUR TEACHER The equipotential surfaces surrounding a point charge are concentric spheres with the charge at the center. If the electric potential of two such equipotential surfaces that surround a point charge +1.43 ✕ 10−8 C are 215 V and 71.5 V, what is the distance between these two surfaces?

Answer 1

Answer:

1.203 m

Explanation:

The expression for electric potential is given as,

V = kq/r

Where V =  Electric potential, k = coulomb's constant, q = charge, r = distance.

From the question,

For the first surface,

V₁ = kq/r₁.................... Equation 1

Where V₁ = Electric potential of the first surface, r₁ = distance of the first surface from the charge.

make r₁ the subject of the equation,

r₁ = kq/V₁................. Equation 2

Given: q = 1.43×10⁻⁸ C, k = 9×10⁹ Nm²/C², V₁ = 215 V

Substitute into equation 2

r₁ =  1.43×10⁻⁸(9×10⁹)/215

r₁ = 128.7/215

r₁ = 0.597 m

For the second surface,

Similarly,

r₂ = kq/V₂................... Equation 4

Given: V₂ = 71.5 V

Substitute into equation 4

r₂ = 1.43×10⁻⁸(9×10⁹)/71.5

r₂ = 128.7/71.5

r₂ = 1.8 m.

Hence the distance between the surface = r₂-r₁

= 1.8-0.597 = 1.203 m