# Calculating Work Required to Move Charge -0.51x10^-12 C

• shimizua
In summary, the equation is W=-q\DeltaV and the attempt at a solution was to find q by using the picture. q is given as -0.51x10^-12 C and the problem is finding \DeltaV without knowing E. The voltage at point i is 3.06x10^-12 J and point b is 6kV.
shimizua

## Homework Statement

Calculate the work required to move a charge of -0.51x10^-12 C from i' to b'.

## Homework Equations

i know that the equation is -W=qEl and that it can also be expressed as -W=-q$$\Delta$$V or W=q$$\Delta$$V

## The Attempt at a Solution

well q is given as -0.51x10^-12
the problem i am having is how do i find $$\Delta$$V without knowing E? cause i know i can figure out l just by using the picture. So if you can just help me with find out how to get $$\Delta$$V that would be great!

Looks like your ΔV is going from +3 to -3. That looks like a Voltage drop of 6v from the image, or ΔV = -6.

Just count the contours is my method.

The figure is a contour plot of the potential. Each contour line represents a constant potential value.

The figure also indicates which contour lines correspond to -5V, 0V, and +5V. Using that, can you say:
What is the voltage at point i?
What is the voltage at point b?

ok so i did the (-.51x10^-12)(-6) and got an answer of 3.06x10^-12 J and it was said to be wrong. Where did i got wrong?

oh and i was given this hint when i put my answer in

The equipotential lines shown are separated by 1 kV. (NOTE! That's kV, not volts!) Work to move a charge is the increase in potential energy of the charge. The potential energy is the potential times the charge.

The answer remains the same though, whether it is kV or V. It's still a matter of counting contour lines.

shimizua said:
ok so i did the (-.51x10^-12)(-6) and got an answer of 3.06x10^-12 J and it was said to be wrong. Where did i got wrong?

If it's actually 1 kV, and not 1 V, for the contour intervals, then instead of 6V it is really ____?

ok so then it would mean that it was really 6 kV or 6000 V and then when i did that i got 3x10^-9 J and that also came up wrong

haha, it is the graph that they gave up. i didnt make it

ok i give up... i am having my own probelms...

It's plus Work moving an opposite charge away from + and toward a - V.

If the contours are 103v then it's 3.06 * 10-9 J or 306 ergs.

If it's not precision or units then I've no idea what is asked.

i got the answer. i did the 3.06e-09 J and it seemed to work even though i did 3.1e-09 before. i guess it had to be 06. thanks for all your help guys

## 1. How do you calculate the work required to move a charge?

To calculate the work required to move a charge, you can use the formula W = qΔV, where W is the work in joules, q is the charge in coulombs, and ΔV is the potential difference in volts.

## 2. What is the unit for work in this context?

The unit for work in this context is joules (J).

## 3. Can you provide an example of calculating work required to move a charge?

Sure, for example, if you have a charge of -0.51x10^-12 C and a potential difference of 10 volts, the work required to move the charge would be W = (-0.51x10^-12 C)(10 V) = -5.1x10^-12 J.

## 4. How does the direction of the charge affect the calculation of work?

The direction of the charge does not affect the calculation of work required to move it. The formula W = qΔV takes into account the sign of the charge, so a negative charge will result in a negative work value.

## 5. What other factors may affect the work required to move a charge?

The work required to move a charge may also be affected by the presence of any external electric or magnetic fields, as well as the distance the charge needs to be moved and the medium it is moving through.

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