 #1
RoboNerd
 410
 11
Homework Statement
Hi, everyone.
I have a square and on each vertex of the square is a point charge of "+q".
The square has a side length of "r" and these four charges form a system.
My problem says to find the work required to assemble the charges shown above, bringing each charge in from infinity is equal to
[k = 1 / (4*pi*e0)
A) 4*k*q^2/r
B) k * (4 + sqrt(2) ) * q^2
C) K * (4 + 2*sqrt(2) ) *q^2 / r
D) K * (8 + sqrt(2) ) *q^2 / r
E) k * (8 + 2*sqrt(2) ) *q^2 /r
E is the right answer in this situation. My problem is figuring out how to solve this
Homework Equations
U = q * change in electric potential
change in electric potential when I am bringing in a charge from infinity to its position is going to be the negative of the Integral of E dot dl, from infinity to its final position.
The Attempt at a Solution
Here's what I tried:
I call the charges in the following positions A, B, C, D for reference.
A  B




DC
I say that the total work done in assembling the system is equal to the work done in assembling each individual charge.
Charge A is the first. It is fixed. WorkDoneForA = 0 Joules.
I bring in charge B next. WorkDoneForB = q*deltaV = k * q^2 / r
Next, I bring in charge c.
WorkDoneForC = Work done in relation to charge A + work done in relation to charge B =
[ k * q^2 / r] + [k *q^2 / (r * sqrt(2) ] = (k*q^2 /r) * ( 1 + 1/sqrt(2) )
WorkDoneForD = Work done in relation to charge A + work done in relation to charge B + work done in relation to charge C
= [ k * q^2 / r] + [k * q^2 / ( r* sqrt(2) ) ] + [ k * q^2 / r] = (k*q^2 / r ) * (2 + (1/sqrt(2)) )
=( k*q^2 / r ) * (4 + 2/sqrt(2) )Thus, as you can see, my answer does not match E, which is the right answer.
Could anyone please advise me as to why this is the case? Thanks in advance, and make it a great day!