Effect of radius changes on electric fields and potential difference?

[mirandab17]

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.

 

[Simon Bridge]

So do the math - calculate the field before and after the change in position.
Note: E is a vector.

 

[mirandab17]

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]

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