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A uniform electric field of magnitude 4.1 ✕ 105 n/c points in the positive x direction. find the change in electric potential energy of a +4.9 µc charge as it moves from the origin to each of the points given below.

Respuesta :

aabdulsamir

Answer:

2.009N

Explanation:

Force on a charge is equal to the product of the charge ans electric field intensity of the charge.

F = qE

q = 4.9E-6C

E = 4.1E5 N/C

F = qE

F = 4.9E-6 * 4.1E5

F = 2.009N