# Calculating Work Required to Move Test Charge

• dotcom
That's why the two terms are added.In summary, the task is to calculate the work required to move a +0.5uC test charge from a midway between two +30uC charges to a point 10cm closer to either of the charges. Using the equations E=V/d, W=qV, and k=8.99*10^9, the correct approach is to find the difference between the initial and final potential energies, which are calculated by taking into account the forces exerted by both charges on the test charge. After making this calculation, the final energy is subtracted from the initial energy to find the work required, resulting in the answer of 0.1498J.

## Homework Statement

A +30uC charge is placed 40cm away from an identical +30uC charge. How much work would be required to move a +0.5uC test charge from a midway between them to a point 10cm clover to either of the charges?

## Homework Equations

E=V/d, W=qV, k=8.99*10^9

## The Attempt at a Solution

well...

W=qV=(0.5*10^(-6)){(30*10^(-6)k/0.1)-(30*10^(-6)k/0.2)}=1.32

Please, anyone who understands this question, explain me how to get the answer!

dotcom said:

W=qV=(0.5*10^(-6)){(30*10^(-6)k/0.1)-(30*10^(-6)k/0.2)}=1.32

There shouldn't be a negative sign there, so that part of your answer is wrong.

I think you need to find the change in potential energy from when the test charge is sitting between the two +30uC charges, to when it is moved 10cm over. That is the work done in moving it. So if you fix the calculation above, you are half done.

I tried with
W=qV=(0.5*10^(-6)){(30*10^(-6)k/0.1)(30*10^(-6)k/0.2)}
but the answer was then too big...

I then assumed that the potential energy is 0 when the test charge is sitting between the two +30uC charges (as it's balanced).
And the energy when it is moved 10cm over is:

force to the left=(0.5*10^(-6))(30*10^(-6))k/0.1^2=13.485
force to the right=(0.5*10^(-6))(30*10^(-6))k/0.3^2=1.498

using E=F/q,
E=(13.485-1.498)/(0.5*10^(-6))=11.987

therefore, W=qEd=0.5*10^(-6)*11.987*0.1=1.198

which is not the right answer...
can anyone suggest me the correct way to go?

The PE of the test charge sitting between the others is just twice what you'd get if there was only one 20cm away.

After moving, you are 10cm from one charge and 30cm from the other. Find the difference between the initial and final energies.

So the initial PE is

net initial force =2*((0.5*10^(-6))(30*10^(-6))k/0.2^2 =6.7425
PE=F/q=6.7425/(0.5*10^(-6))=13485000

W=qEd=(0.5*10^(-6))*13485000*0.2=1.3485

net final force =((0.5*10^(-6))(30*10^(-6))k/0.1^2 +((0.5*10^(-6))(30*10^(-6))k/0.3^2=14.9833
PE=F/q=14.9833/(0.5*10^(-6))=29966666

W=qEd=(0.5*10^(-6))*29966666*0.1=

W required=1.4983-1.3485=0.1498

I still cannot get the right answer...
Where did I fall into a trap?

Well, you can't be faulted for effort. You're doing it the hard way.
You don't need to calculate the forces, because PE = k.q.q/r.

So,

initial energy=2*((0.5*10^(-6))(30*10^(-6))k/0.2 = ??
final energy=((0.5*10^(-6))(30*10^(-6))k/0.1 +((0.5*10^(-6))(30*10^(-6))k/0.3=??

and

work = final energy - initial energy

Last edited:
Oh, I finally got the right answer, following Mr/Ms Mentz114's way! Thank you so much!

But...I still can't understand why the energy in both directions are added together for calculating the potential energy. Aren't the charges at right and left end both positive? Then don't they exert equal and opposite force on the test charge in the midway?

Your question is a good one. If we moved the test charge to the left, the big charge on the right is helping us, so we are doing negative work wrt to it. But the big charge on the right opposes us, so we do positive work against it.
There are two terms in the final energy, and if we compare with the two (equal) terms in the initial energy, one as gone up, and the other down.

## 1. What is the definition of "work" in physics?

In physics, work is defined as the amount of energy transferred to or from an object by means of a force acting on the object. It is typically measured in joules (J).

## 2. How do you calculate the work required to move a test charge in an electric field?

The work required to move a test charge in an electric field can be calculated by multiplying the magnitude of the electric field by the distance the charge is moved in the field. This can be expressed as W = qEd, where W is the work, q is the test charge, E is the electric field, and d is the distance moved.

## 3. What is the unit of measurement for work in the context of moving a test charge?

The unit of measurement for work in the context of moving a test charge is the joule (J).

## 4. How can you determine the direction of the work required to move a test charge?

The direction of the work required to move a test charge is determined by the direction of the electric field and the direction in which the charge is being moved. If the charge is moving in the same direction as the electric field, the work will be positive. If the charge is moving in the opposite direction of the electric field, the work will be negative.

## 5. Can the work required to move a test charge be negative?

Yes, the work required to move a test charge can be negative if the charge is moving in the opposite direction of the electric field. This means that energy is being transferred from the charge to the field, rather than from the field to the charge.