Calculate the electric potential difference

+4 votes

A uniform electric field of magnitude 290 V/m is directed in the negative y direction as shown in the figure below. The coordinates of point circle a are (-0.600, -0.500) m, and those of point circle b are (0.450, 0.350) m. Calculate the electric potential difference

VBVA using the dashed-line path.


asked Jan 20, 2012 in Physics by Roar3215 ~Expert~ (1,318 points)

1 Answer

+3 votes
Best answer

The formula for change in electric potential is:

VB - VA = -(integral from point A to point B  of  (E * ds)), where E is the electric field.

E =  290 V/m as given in question.

You can solve this integral by breaking it up into 2 seperate integrals.  To do so, lets call the corner of where the 2 points meet, point C, which will have coordinates (-0.60, 0.35)

So break it up into an integral from A to C, and one from C to B as follows:

- integral from A to C (E*dy)*cos180, replace A and C with the respective y coordinates to get:

- integral from -0.50 to 0.35 (E*dy) * cos180

*note the cos 180 is the angle between the between the charge going in the direction from A to C, and that of the electric field, since they are going in opposite directions the angle is 180.

So simply solve the intergral and plug in E to solve,

- E*cos180 * (0.35 - (-.50)) =

(290 V/m)*(0.85)m = 1246.5 V


and the second integral is:

- integral from C to B (E*dx)*cos90, replace B and C with the respective x coordinates to get:

- integral from -0.6 to 0.45 (E*dx)*cos90

*the angle of 90, here comes from the the fact that the charge from C to B is at a 90 degree angle from that of the electric field. 

Cos90 = 0, and so the second integral is simply 0, which makes sense considering that electric potential does not change if moving horizontally with respect to direction of the electric field

Therefore the final answer is just: 1246.5 V



answered Jan 24, 2012 by pokemonmaster ~Expert~ (3,854 points)
selected Jan 24, 2012 by Roar3215

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