Electric Potential Difference b/w A & B: Solving for VA - VB

[tag16]

Homework Statement

Two points (A and B) are shown in a constant electric field of E = 850 N/C. The distance between A and B is L = 2 m. The line joining the two points makes an angle of 40 degrees with the electric field. Determine the electric potential difference (in Volts) between points A and B -- that is VA - VB.

Homework Equations

E_s cos(theta)=-Ed

The Attempt at a Solution

would I just do E(cos theta) or -Ed? or are neither right?

 

[Delphi51]

The way I like to think of it is that the change in potential V is the work that must be done per unit charge to move a charge from A to B. The work is the force that must be exerted times the distance for which it is exerted. So we have
V = W/q and W = Fd or V = Fd/q
The electric field causes a force on the charge we are moving, F = qE.
However this force is not in the exact direction we are pushing the charge. So the force we need to overcome is actually qE*cos(A) where A is the angle between the E field and the direction we are going. That leaves us with a potential difference of
V = Fd/q = qE*cos(A)*d/q.

 

[tag16]

except that q is not given in the problem

 

[Delphi51]

The q on the top cancels with the q on the bottom.

 

[tag16]

yeah I thought that might be it so I tried it that way but still got the wrong answer

 

[Delphi51]

I'm not supposed to show you the calculation for fear of spoiling the experience for you. But if you show your calc, we can check it.