Distance between equipotential surfaces

AI Thread Summary
The discussion revolves around calculating the distance between equipotential surfaces for a non-conducting sphere with a uniform charge density. The initial calculations led to incorrect values for the radii corresponding to the given potentials, raising questions about unit consistency and algebraic errors. After re-evaluating the calculations, the corrected distances were found to be approximately 1.998 meters for V1 and 0.7651 meters for V2, indicating that the radius decreases with increasing potential. The final distance between the equipotential surfaces was determined to be 1.23 meters. The conversation highlights the importance of careful unit management and algebraic accuracy in physics problems.
Alan I
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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. :oldconfused:
 
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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.
 
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mfb said:
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!
 
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