# Work done in assembling a system of charges

• RoboNerd
In summary, the conversation discussed finding the work required to assemble four point charges on the vertices of a square with side length "r". The correct answer was determined to be option E, which is equal to k * (8 + 2*sqrt(2) ) * q^2 / r. The conversation also mentioned a possible mistake in counting contributions twice and suggested simplifying 2/sqrt(2).
RoboNerd

## 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
|
|
|
|
D------------------------------C
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!

You would get (E) if you mistakenly count every contribution twice.

You can simplify 2/sqrt(2), by the way.

RoboNerd

## 1. What is work done in assembling a system of charges?

The work done in assembling a system of charges is the amount of energy required to bring the charges from an infinite distance and arrange them in their final positions. It is also known as the potential energy of a system of charges.

## 2. How is work done in assembling a system of charges calculated?

The work done in assembling a system of charges can be calculated by using the formula W = qV, where W is the work done, q is the charge of each individual charge, and V is the potential difference between the initial and final positions of the charges.

## 3. Does the work done in assembling a system of charges depend on the order in which the charges are brought together?

No, the work done in assembling a system of charges is independent of the order in which the charges are brought together. This is because the potential energy of a system of charges only depends on the final positions of the charges, not the path taken to reach those positions.

## 4. How does the distance between charges affect the work done in assembling a system of charges?

The work done in assembling a system of charges is directly proportional to the distance between the charges. This means that as the distance between charges increases, the work done also increases. This is because the potential energy of a system of charges decreases as the distance between them increases.

## 5. Can the work done in assembling a system of charges be negative?

Yes, the work done in assembling a system of charges can be negative. This occurs when the charges are brought closer together, resulting in a decrease in potential energy. In this case, work is actually being done by the system on the charges, rather than the other way around.

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