How Much Work Is Needed to Position a Charge in an Equilateral Triangle?

[indie452]

Homework Statement

situation:
there are two charges (+q) one at one corner of an equilateral triangle the other at another corner. the triangle has sides length a.

what is the work required to bring another charge (+q) in from infinity to the other corner on the equilateral triangle.

The Attempt at a Solution

ok so i know dW = -dU = F.dl

i thought that maybe finding (F) at the corner that we are bringing the charge to may help...this is:
F_c [at corner] = q² / 2*pi*epsilon*a²

then i could just integrate this over the distance i am moving it with respect to a...however i know that the one limit will be $$\infty$$ however i don't know what the other will be

thanks for any help

 

[LowlyPion]

Consider the Voltage at the point from each of the 2 other charges.

V = k*q / r

Since work = q*ΔV

and V_∞ = 0

Then

Work = q*ΣV

 

[indie452]

ok so
the E at the point is 2q / 4*pi*epsilon*a²
the V at the point is 2q / 4*pi*epsilon*a

so work to bring in point is = q(2q / 4*pi*epsilon*a) = q² / 2*pi*epsilon*a

 

[LowlyPion]

ok so
the E at the point is 2q / 4*pi*epsilon*a²
the V at the point is 2q / 4*pi*epsilon*a

so work to bring in point is = q(2q / 4*pi*epsilon*a) = q² / 2*pi*epsilon*a

Looks like it.

Though E is a vector, don't forget, and V is a scalar here. In your first equation then the E would need to be added as vectors. Whereas for V you are adding scalars.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html#c1

 

[indie452]

ok thanks...