# Homework Help: Distance between equipotential surfaces

1. Sep 16, 2015

### Alan I

1. The problem statement, all variables and given/known data
A non-conducting sphere (radius 11.3 cm) has uniform charge density ρ = 0.596 μC/m3. Find the distance, in meters, between equipotential surfaces V1 = 16.2 Volts and V2 = 42.3 volts. (Distance is always positive.)

2. Relevant equations
V=kq/r
ρ=Q/V

3. The attempt at a solution
ρ=0.596*10-6 C/m3 = Q/V ⇒ (0.596*10-6)=Q/(4/3*π*(0.113m)3)
⇒Q=3.60*10-9

for V1
16.2=k(3.60*10-9)/r1
⇒r1≅0.5

for V2
42.3=k(3.60*10-9)/r2
⇒r2≅1.3

⇒r2-r1=0.807m → this answer is wrong.

2. Sep 16, 2015

### Staff: Mentor

Volts relative to what?

There are missing units.
How do you get a larger radius for the larger potential? Something got wrong with the multiplication/division there, you got the inverse values. An error that would have been easy to spot with units.

3. Sep 16, 2015

### Alan I

OK thanks! I don't know how I messed up that algebra so bad. After re-checking my calculations this is what I got:

r1 m = k N*m2/C2 * (3.6*10-9) C / 16.2 N*m/C

r1 = 1.998 m

r2 m = k N*m2/C2 * (3.6*10-9) C / 42.3 N*m/C

r2=0.7651 m → smaller for the higher potential

r1-r2 = 1.23 m

Now it seems to make more sense. Thank you!