Question

Two metal spheres have a center to center separation of d=2m. Sphere 1 has radius R1=.03m and charge q1=+1*x10^-8 C. Sphere 2 has radius R2=2xR1 and charge q2=-3*x10^-8 C. Assume the charge on each sphere is uniformly distributed. With V=0 at infinity, calculate the potential on the surfaces of Sphere 1 and Sphere 2.

Answer 1

Answer:

V₁ =2865 V

V₂ = -4455 V.

Explanation:

Given:

Charge on sphere 1 = q₁ = 1 × 10⁻⁸ C

Charge on the sphere 2 = q₂ = -3 ×10⁻⁸ C

Radius of sphere 1 = R₁ = 0.03 m

Radius of sphere 2 = R₂ = 2 R₁ =0.06 m

Distance between the two spheres = d = 2 m

Potential on the sphere 1 = V₁ = $k[\frac{q_{1}}{R_{1}} + \frac{q_{2}}{d}]$ where k is the Coulomb's constant.

⇒ V₁ = $(9\times 10^{9})[\frac{1\times 10^{-8}}{0.03}+(\frac{-3\times 10^{-8}}{2})]$ = 2865 V

Similarly, Potential on V₂ = $(9\times 10^{9})[\frac{1\times 10^{-8}}{2}+(\frac{-3\times 10^{-8}}{0.06})]$ = -4455 V.