Radii of eqiuipotential surfaces of a charged metal sphere

AI Thread Summary
A metal sphere with a radius of 0.29 m and a charge of 0.90 µC requires the calculation of equipotential surfaces at 100 V intervals. The formula V = kQ/r indicates that the radius r must account for the sphere's radius, as the potential at the surface is significantly high (approximately 27931.03 V). Attempts to calculate the first, tenth, and hundredth equipotential surfaces by simply adding 100 V increments to the surface potential yielded incorrect results. The correct approach involves adjusting the radius calculations to include the sphere's radius when determining the distance from its surface. Understanding the relationship between potential and distance is crucial for accurate calculations.
RoosterCancer
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Homework Statement



A metal sphere of radius r0 = 0.29 m carries a charge Q = 0.90 µC. Equipotential surfaces are to be drawn for 100 V intervals outside the sphere. Determine the radius r of the following equipotentials from the surface.
a) first:___m
b) tenth:___m
c) 100th:_____m


Homework Equations



V=kQ/r --> r=kQ/V



The Attempt at a Solution



I've tried plugging the numbers from the problem in the equation:

r=(8.99 x 10^9 NM^2/C^2)*(0.9 x 10^-6 C)/(100V) = 80.91m

and it's incorrect. I assume the radius of the sphere itself comes into play, but I'm not sure where. Can anyone help?
 
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RoosterCancer said:

Homework Statement



A metal sphere of radius r0 = 0.29 m carries a charge Q = 0.90 µC. Equipotential surfaces are to be drawn for 100 V intervals outside the sphere. Determine the radius r of the following equipotentials from the surface.
a) first:___m
b) tenth:___m
c) 100th:_____m

Homework Equations



V=kQ/r --> r=kQ/V

The Attempt at a Solution



I've tried plugging the numbers from the problem in the equation:

r=(8.99 x 10^9 NM^2/C^2)*(0.9 x 10^-6 C)/(100V) = 80.91m

and it's incorrect. I assume the radius of the sphere itself comes into play, but I'm not sure where. Can anyone help?
What is the potential at the surface of the ball ?
 
I was able to find the potential of the ball:

V=(9 x10^9NM^2/C^2)*(0.9 x10^-6 C)/0.29m = 27931.03V

and was able to answer part a, by adding 100 V to the ball's potential and using the formula. So, for b, I added 1000 V to the potential and for c I added 10000 V to the potential, but the results were incorrect.
 
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