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 1.63 10-8 C are 200 V and 82.0 V, what is the distance between these two surfaces?

1.06 m

Explanation:

Since the charge is at the centre of two concentric spheres, we use the formula for electric potential due to a point charge. V = kq/r. Let r₁ be the radius of the sphere with potential, V₁ = 200 V and  r₂ be the radius of the sphere with potential, V₂ = 82.0 V. From V = kq/r, r = kq/V. So that r₁ = kq/V₁ and r₂ = kq/V₂. The magnitude of the difference r₁ - r₂ is the distance between the two surfaces. q the charge equals 1.63 × 10⁻⁸ C

r₂ - r₁ = kq/V₂ - kq/V₁ = kq(1/V₂ - 1/V₁) = 1.63 × 10⁻⁸ × 9 × 10⁹ (1/82 -1/200) m =  1.63 × 10⁻⁸ × 9 × 10⁹ (0.0122 - 0.005) = 1.63 × 10⁻⁸ × 9 × 10⁹(0.0072) m = 1.06 m

The distance between them is 1.06 m

c)as speed changes, wavelength and frequency change.

everything changes!

the answer is a. hope this .