Method of Images Problem

[blizzardof96]

Homework Statement

The z = 0 plane is a grounded conducting surface. A point charge q is at (0,0,a), and charge 4q at (0,-2a,a).
Calculate the potential in the region z > 0.

Homework Equations

V=∑kq/r

The Attempt at a Solution

[/B]
Use the method of images.

V1 = kq/r+ + kq/r-
V1=kq(1/sqrt(x^2 +y^2 + (z-a)^2) - 1/sqrt(x^2 +y^2 + (z+a)^2)

V2=4kq/r+ + 4kq/r-
V2 = 4kq(1/sqrt(x^2 + (y+2a)^2 + (z-a)^2) - 1/sqrt(x^2+(y+2a)^2 + (z+a)^2))

Vtotal=ΣV=V1+V2
Vtotal=4kq(1/sqrt(x^2 + (y+2a)^2 + (z-a)^2) - 1/sqrt(x^2+(y+2a)^2 + (z+a)^2)) + kq(1/sqrt(x^2 +y^2 + (z-a)^2) - 1/sqrt(x^2 +y^2 + (z+a)^2)

Is this the correct answer?

 

[BvU]

PF isn't for stamp-approving homework. The answer to your question can be obtained from your teacher: hand it in.
Or is there another reason you ask: have doubt at some point ? Tell us.

 

[blizzardof96]

PF isn't for stamp-approving homework. The answer to your question can be obtained from your teacher: hand it in.
Or is there another reason you ask: have doubt at some point ? Tell us.

I have doubts as the value for Vtotal seems unnecessarily complex and there is nothing in the final expression that can simplify. The solution appears much different from the usual form we see in an electric potential problem.

 

[BvU]

unnecessarily complex

I see. Lots of textbook exercises have compact and elegant answers, I agree. Somewhat unrealistic, I must warn you.

In this case you have four terms from the four contributing charges (two real, two image). The mirror image method ensures the physical boundary conditions are met ( V=0 at z=0 and $\vec E \perp (z=0)$ ) and that's about all there is to be said.

At large distances ( $|\vec r| >> a$ ) your configuration should give a dipole field, but here you are clearly expected to give the full detailed answer you worked out.

 

[blizzardof96]

I see. Lots of textbook exercises have compact and elegant answers, I agree. Somewhat unrealistic, I must warn you.

In this case you have four terms from the four contributing charges (two real, two image). The mirror image method ensures the physical boundary conditions are met ( V=0 at z=0 and $\vec E \perp (z=0)$ ) and that's about all there is to be said.

At large distances ( $|\vec r| >> a$ ) your configuration should give a dipole field, but here you are clearly expected to give the full detailed answer you worked out.

Thank you. As a third year undergrad with less physics knowledge than yourself I appreciate it.