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A uniform electric field with a magnitude of 6860 N/C points in the positive x direction.

A.) Find the change in electric potential energy when a +19.5-μC charge is moved 4.25 cm in the positive x direction.
B.) Find the change in electric potential energy when a +19.5-μC charge is moved 4.25 cm in the negative x direction.
C.) Find the change in electric potential energy when a +19.5-μC charge is moved 4.25 cm in the positive y direction

User Jeremias
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Final answer:

The change in electric potential energy when a charge is moved through a uniform electric field is calculated based on the charge's displacement in the direction of the field. It's positive when moving with the field, negative when against it, and zero when perpendicular to the field.

Step-by-step explanation:

The change in electric potential energy (U) in a uniform electric field (E) can be determined using the formula U = qEd, where q is the charge and d is the displacement in the direction of the electric field. Given an electric field magnitude of 6860 N/C, a charge of +19.5-μC, and a displacement of 4.25 cm (converted to meters), we can calculate the change in electric potential energy in each scenario as follows:

  • A) When the charge is moved in the direction of the electric field: U = (19.5 × 10-6 C) × (6860 N/C) × (0.0425 m) = 5.679075 × 10-3 J
  • B) When the charge is moved against the direction of the electric field: U = -(19.5 × 10-6 C) × (6860 N/C) × (0.0425 m) = -5.679075 × 10-3 J
  • C) When the charge is moved perpendicular to the electric field (in the positive y direction): U = 0 because the displacement is perpendicular to the electric field direction, so no work is done against the electric field.

User Mtjhax
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