Effect of radius changes on electric fields and potential difference?


by mirandab17
Tags: electric field, electric potential
mirandab17
mirandab17 is offline
#1
Jan31-12, 01:21 AM
P: 40


Hello!

Okay so I understand that electric potential:

V = kQ/r

...must be influenced by the radius doubling because it would make the potential energy half of what it originally was because of the proportionality law, v is proportional to 1/r.

With electric fields though, how can there possibly be no change? The formula is

E = kQ/r^2

...meaning if should be influenced as well?

The answer is c, btw.
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Simon Bridge
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#2
Jan31-12, 05:39 AM
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So do the math - calculate the field before and after the change in position.
Note: E is a vector.
mirandab17
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#3
Jan31-12, 08:46 AM
P: 40
Oh! Right!

Since E is a vector, and the distance and charges on both sides are equal, then they always simply cancel out to zero. Whereas with electric potential, a scalar quantity, it is not affected by direction, merely magnitude, in which case both are positive, so the radius change will definitely affect it.

Thanks bud!

Simon Bridge
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#4
Jan31-12, 09:14 AM
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Effect of radius changes on electric fields and potential difference?


No worries - that "Oh! Right!" feeling is what I was going for :)


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