electrostatics question

[swamprat]

Q:

When placed a distance d from a + point charge a positively charged particle has a PE of U due to the electric field created by the point charge. If the charge is moved to a distance of 2d, which of the following represents its PE?

Answer is U/2 but I got U/4.

Isn't the equation for electric field due to a point charge:

E= kq / r^2

So if r is doubled, you square the 2 and that gives you 4. It was must understanding that the only electrostatic equation that does not square the distance is:

V = kq / r

Which I thought woudn't apply here.

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[swamprat]

Act now that I think about it, E = N/C and PE = J ... so could someone point me in the right direction?

 

[thatscorrect7]

Force x distance

 

[drillr00]

This is my understanding.

Definition of Electric Potential: U=W=qV

When there's only one other charge: V= kQ/r

Substituting in for V into the Electric Potential Equation you get: U= k(qQ)/r

So doubling the distance makes U now U/2

 

[marele86]

You're right that Voltage falls off as 1/r, while Force and Electric field both fall off as 1/r^2, but U (potential energy) is very similar to voltage.

It's just like electric field is F/q (i.e., an electric field measures the 'force' just from one of the charges, not both - so we call it a field):

Voltage is just a measure of U/q. Same thing. Take the potential energy, but then divide it by the second charge so we're just seeing the electric potential of the first charge (i.e. How much work does it take PER CHARGE to approach that charge?)

Therefore Voltage really only falls off as 1/r because potential energy falls off as 1/r.