Distance between equipotential surfaces

[Alan I]

Homework Statement

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.)

Homework Equations

V=kq/r
ρ=Q/V

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.

 

[mfb]

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.

 

[Alan I]

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.

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!