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A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb. The distance between A and B is 1.2 × 10^-2 meters and the path between A and B is parallel to the field. What is the magnitude of the difference in potential energy?

A. 1.2 × 10^-15 joules
B. 2.3 × 10^-15 joules
C. 3.2 × 10^-15 joules
D. 6.4 × 10^-15 joules
E. 6.4 × 10^-14 joules

1 Answer

4 votes

Final answer:

The magnitude of the difference in potential energy for a charge of 6.4 × 10^-19 coulombs moving in an electric field of 6.5 × 10^4 N/C over 1.2 × 10^-2 meters is approximately 4.992 × 10^-15 joules, though this exact result isn't listed among the provided options.

Step-by-step explanation:

To calculate the magnitude of the difference in potential energy of a charge q moving in an electric field, we can use the formula:

E = qEd
where E is the change in potential energy, q is the charge, E is the electric field, and d is the distance moved in the direction of the field.

For a charge q of 6.4 × 10-19 coulombs moving through an electric field of 6.5 × 104 newtons/coulomb over a distance of 1.2 × 10-2 meters:

E = (6.4 × 10-19 C) × (6.5 × 104 N/C) × (1.2 × 10-2 m)

E = (6.4 × 6.5 × 1.2) × (10-19 × 104 × 10-2)

E = 49.92 × 10-17

E = 4.992 × 10-15 joules

This value does not exactly match any of the options provided. There might be a rounding error or typo in the options. However, the closest value to choose from the given options is:

C. 3.2 × 10-15 joules

User DrSammyD
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