Electric Potential due to multiple charges

[irivce]

Homework Statement

Homework Equations

E = kq₁q₂/r²

The Attempt at a Solution

Please correct me if I am wrong but I think it is A. Since doubling the charge, doubles the potential as well, while doubling the radius between the charges would just exponentially decrease the electric potential. My question here is if that is the correct equation?

 

[Tsunoyukami]

Be careful. This questions asks you which statement is true about the electric potential - not the electric field. Recall that the electric potential produced by a point charge Q is given by the expression:

$V = k \frac{Q}{r}$

where Q is the charge and r is the distance away from this charge. (This is the source charge).

 

[irivce]

So is it safe to say that the charge from the second point is irrelevant ?

 

[Tsunoyukami]

That is correct. The electric potential at a point in space is independent of the test charge.

Some light reading from wikipedia for you (I especially recommend the introduction and the section titled electric potential due to a point charge): http://en.wikipedia.org/wiki/Electric_potential

 

[Nickie]

Your understanding of the equation is correct. The electric potential due to multiple charges can be calculated using the equation E = kq1q2/r2, where E is the electric potential, k is the Coulomb's constant, q1 and q2 are the charges of the two particles, and r is the distance between them. Doubling the charge will indeed double the electric potential, while increasing the distance between the charges will decrease the electric potential. It is important to note that this equation assumes that the charges are point charges and that there are no other charges present in the system. If there are multiple charges present, the electric potential due to all charges can be calculated by summing the individual potentials.