How to picture a radial field around a 3d object?

AI Thread Summary
The discussion focuses on visualizing a radial field around a spherical conductor, comparing it to an atom and addressing the impact of diameter on field representation. It clarifies that the surface voltage of 15,000 V relates to equipotential lines that decrease according to a 1/r relationship. A participant expresses confusion over their calculations of electric field strength, which differed from the expected value. The correct approach involves treating the spherical conductor as a point charge and using the radius for potential calculations. The conversation concludes that as long as the charge distribution remains symmetric, the spherical object can be effectively analyzed as a point charge.
jackiepollock
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Homework Statement
The problem is asking me to imagine a radial field around a Van de Graaff generator and finding the potential values and electric field strength in varying distances from the surface.
Relevant Equations
For electric field strength:E= Q/4πϵ0r^2

For potential :E= Q/4πϵ0r
Hello!

First off, for a), I am not too sure how to picture a radial field around a 3d object. I know that this spherical metal dome is basically a enlarged version of an atom, but since with problems on radial field around an atom, I don't have to consider its diameter, I'm not sure how the diameter of 30cm should make a difference to picturing the radial field.

for b),
The answer says: equipotentials concentric with dome surface; increasing separation of equipotentials; surface voltage is 15 000 V, so equipotential labelling should decrease from that following 1/r relationship. I am not too sure how this 15000V is calculated.

for c)
I tried 8.99 x 10^9 x 2.5 x 10^-7/(0.05)^2, which is 899000 Vm^-1, different from the answer provided, being 56000 Vm^-1. What did I do wrong?

Thank you so much for the help!

A Van de Graaff generator is charged with 2.5 x 10-7 coulombs in a.png
 
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First: don't use the same symbol for field and potential!
The spherical conductor can be treated (from outside) as if it was a point charge located at the centre of the sphere. So for the surface voltage, use the potential equation with r = the radius of the sphere.
What is r at a distance of 5 cm from the surface?
 
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Likes Delta2, DaveC426913 and jackiepollock
mjc123 said:
First: don't use the same symbol for field and potential!
The spherical conductor can be treated (from outside) as if it was a point charge located at the centre of the sphere. So for the surface voltage, use the potential equation with r = the radius of the sphere.
What is r at a distance of 5 cm from the surface?
Got it! I've tried what you said and gotten the right answers. Thank you so much!
 
mjc123 said:
The spherical conductor can be treated (from outside) as if it was a point charge located at the centre of the sphere.
There is a caveat that can be important. A spherical object can safely be treated as a point charge if its charge distribution is spherically symmetric. A spherical conductor will relax to have a uniform charge distribution if there is no external field causing an asymmetry.

One reasonably assumes that no very large asymmetric external field is present in this case. So no significant asymmetry in charge is to be expected. So the caveat is unimportant here.
 
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