# How Is Work Calculated in Electric Fields?

• rojasharma
In summary: Essentially, you would be finding the work done in moving the charges from their initial position (r1) to their final position (r2).
rojasharma
a test charge of +1.0x10^-6C is 40cm from a charged sphere of 3.2x10^-3C. A) how much work was required to move it therer from a point 1.0x10^2m away from the sphere? b) how many electrons were gained or lost from the test object to creat the charge?. I tried but i do not get the right answer:(. what i did was..used Ee1=kq1q2/r1 (r1 as 0.4m), Ee2=kq1q2/r2(r2 as 100m), then found delta Ee= Ee2=Ee1, Delta Ee= W??

Work is defined as F*ds. Look at the equations you used and the units they give.

In the book there is excatly same question except it is moved 100cm instead of 100m. And the answer they got was 43 J. i tired all possible ways...i don;t get the right answer. if i use the above equation, delta Ee=E2-E1, it gives me 43..( but i don;t think that's the right approch here:S)

I think you need another equation to solve for this. First, try to find that problem on cramster.com (they have solutions for many problems). We are also learning Electric Fields(physics 11) in my class at the moment.

Don;t see that problem in cramster.com:(, thanks for the web though.

rojasharma said:
what i did was..used Ee1=kq1q2/r1 (r1 as 0.4m), Ee2=kq1q2/r2(r2 as 100m), then found delta Ee= Ee2=Ee1, Delta Ee= W??
That sounds right. You need to find the change in electric potential energy as the charges are brought together.

So,
q1=3.2x10^-3C
q2=1x10^-6C
r1=0.4m
r2=100m

Solution
E1= 72J (E1= (kq1q2)/r1)
E2= 0.288J (E2=kq1q2/r2)

delta E= E2-E1
delta E= 0.288J - 180J
delta E=-179.712J
Delta E= work done...(what about the negative value??)

rojasharma said:
So,
q1=3.2x10^-3C
q2=1x10^-6C
r1=0.4m
r2=100m

Solution
E1= 72J (E1= (kq1q2)/r1)
E2= 0.288J (E2=kq1q2/r2)
Looks OK. (You seemed to have switched r1 & r2: the charge moves from 100m to 0.4m.)

delta E= E2-E1
delta E= 0.288J - 180J
delta E=-179.712J
Delta E= work done...(what about the negative value??)
Where did you get 180J from? The negative sign is due to you mixing up initial and final positions.

Oh, sorry the E1 is 72J...so its delta E= E1- E2?/ how did i mix up?

Initial position: r1 = 100 m.
Final position: r2 = 0.4 m.

right ...silly me...thanks a lot

Doc Al said:
That sounds right. You need to find the change in electric potential energy as the charges are brought together.

hmm. I am thinking of writing the force between the two charges and integrating it over the distance.

PaintballerCA said:
I am thinking of writing the force between the two charges and integrating it over the distance.
Which is how one would derive the expression for electric potential energy.

## What is work done to move a charge?

The work done to move a charge is the amount of energy required to move a charged particle from one point to another point in an electric field.

## What is the unit of measurement for work done to move a charge?

The unit of measurement for work done to move a charge is joules (J).

## How is work calculated for moving a charge?

Work done to move a charge is calculated by multiplying the magnitude of the charge (in coulombs, C) by the potential difference (in volts, V) through which it moves. The equation for work is W = QV.

## Does the direction of the charge's movement affect the work done?

Yes, the direction of the charge's movement does affect the work done. Work is a scalar quantity, so it is only affected by the magnitude of the charge and potential difference. However, the direction of the charge's movement can affect the sign of the work, indicating whether work is being done on or by the charge.

## What is the relationship between work done and potential difference?

The relationship between work done and potential difference is directly proportional. This means that as the potential difference increases, the work done to move a charge also increases. Similarly, as the potential difference decreases, the work done to move a charge decreases.

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