Question

A metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere❝s radius is 52.5cm and it carries a total charge of +1.45μC .

A) Calculate the potential of the sphere❝s surface.(with the appropriate)

B) You want to draw equipotential surfaces at intervals of 500 V

outside the sphere❝s surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20th and 21st equipotential surfaces.

Answer 1

V(Electric Potential) = U(Potential Energy) / q' (test charge)

since ...

U = (k(constant) x q(source charge) x q'(test charge))/ r (radius)

that means

V = (k x q)/r
V = (8.99 x 10^9) x (1.45 x 10^-6)/ (.52)
V = 25068.67

V = 2.51 x 10^4 volts

Part B:

1st Eq = Surface Electric Potential - (# of intervals x 500 V)
1st Eq = 25068- 500
1st Eq = 24568 V

2nd Eq = Surface Electric Potential - (# of intervals x 500 V)
2nd Eq = 24068 - 1000
2nd Eq = 24068 V

1st Eq = (k x q)/r
24568 = (8.99 x 10^9) x (1.45 x 10^-6)/ (r)
r1 = 0.531

2nd Eq = (k x q)/r
24068= (8.99 x 10^9) x (1.45 x 10^-6)/ (r)
r2 = 0.542

r2 - r1 = ?r
?r = .011

20th Eq = Surface Electric Potential - (# of intervals x 500 V)
20th Eq = 25068- 10000
20th Eq = 15068

21st Eq = Surface Electric Potential - (# of intervals x 500 V)
21st Eq = 25068- 10500
21st Eq = 14568

20th Eq = (k x q)/r
15068= (8.99 x 10^9) x (1.45 x 10^-6)/ (r)
r1 = 0.865

21st Eq = (k x q)/r
14568= (8.99 x 10^9) x (1.45 x 10^-6)/ (r)
r2 = 0.895

r2 - r1 = ?r
?r = .0300