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
