# How much work is done by a system in which the force is 2.8X10^4 N, the potential difference is -11 V and the electric field intensity is 3.9X10^-3 N/C?This question confuses me because it seems to...

How much work is done by a system in which the force is 2.8X10^4 N, the potential difference is -11 V and the electric field intensity is 3.9X10^-3 N/C?

This question confuses me because it seems to me as if there are so many equations to integrate into one another? Any help on how to geanswered answer? Or how to start it out?

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### 1 Answer

In order to determine how much work is done by the system we require two equations. The first equation is the equation for work:

W = F * d

Work is the amount of force applied over a certain distance. The unknown in this equation is the distance over which the known force was applied.

In order to solve for d we call on a second equation which describes the relationship between the potential difference (deltaV), the electric field intensity (E), and the distance (d):

deltaV = -E * d

Substituting in the values we have we get:

11 = -(3.9x10^-3)*d

Rearranging for d:

d = -11/-3.9x10^-3

= 2820 m

Substituting what we have back into our first equation:

W = (2.8x10^4)(2820) = 7.9x10^7 Nm

Therefore, 7.9x10^7 Nm of work is done by the system.

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