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

[shimizua]

Homework Statement

Calculate the work required to move a charge of -0.51x10^-12 C from `i' to `b'.
http://capa-new.colorado.edu/msuphysicslib/Graphics/Gtype54/prob04a_threeqcontour.gif

Homework Equations

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

The Attempt at a Solution

well q is given as -0.51x10^-12
the problem i am having is how do i find \DeltaV 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 \DeltaV that would be great!

 

[LowlyPion]

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.

 

[Redbelly98]

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?

 

[shimizua]

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?

 

[shimizua]

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.

 

[cjl]

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

 

[Redbelly98]

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 ____?

 

[shimizua]

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

 

[Newton V]

your graph is ill-made...

 

[shimizua]

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

 

[Newton V]

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

 

[LowlyPion]

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.

 

[shimizua]

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