A point charge q = +43.0 µC moves from A to B separated by a distance d = 0.182

Question

A point charge q = +43.0 µC moves from A to B separated by a distance d = 0.182 m in the presence of an external electric field Earrowbold of magnitude 281 N/C directed toward the right as in the following figure.

(a) Find the electric force exerted on the charge (both magnitude and direction)

(b) Find the work done by the electric force.

(c) Find the change in the electric potential energy of the charge.

(d) Find the potential difference between A and B.
VB - VA = V

A point charge q = +43.0 C moves from A to B separated by a distance d = 0.182 m in the presence of an external electric field Earrowbold of magnitude 281 N/C directed toward the right as in the following figure. (a) Find the electric force exerted on the charge (both magnitude and direction) (b) Find the work done by the electric force. (c) Find the change in the electric potential energy of the charge. (d) Find the potential difference between A and B. VB - VA = V

Explanation / Answer

Part A)

The formula for force exerted on a charge in an electric field is

F = qE

By substitution, F = (43 X 106)(281) = 1.21 X 1010 N and since it is a positive charge the force will act in the direction of the Electric Field (in this case toward the positive x direction)

Part B)

The formula for work done on a charge is W = qEd

By substitution W = (43 X 106)(281)(.182) = 2.20 X 109 J

Part C)

By definition, W = -change in PE

Therefore the change in potential energy is -2.20 X 109 J

Part D)

The potential difference is defined as the change in Potential Energy divided by the charge

Since the change in PE is -2.20 X 109 J and the charge is 43 X 106 C, simply divide

(-2.20 X 109 J) / (43 X 106 C) = 51.2 J/C or 51.2 V

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