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What is the magnitude and direction of the force exerted on a 3.50μC charge by a 250 N/C electric field that points due east?

a) ( 8.75 X 10³ N ) eastward
b) ( 1.25 X 10² N ) eastward
c) ( 8.75 X 10³ N ) westward
d) ( 1.25 X 10² N ) westward

User Labotsirc
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1 Answer

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Final answer:

The magnitude of the force exerted on a 3.50 μC charge by a 250 N/C electric field that points due east is (8.75 × 10³ N) eastward.

Step-by-step explanation:

The magnitude of the force exerted on a charge can be calculated using the formula:

F = qE

Where F is the force, q is the charge, and E is the electric field strength.
In this case, the charge is 3.50 μC and the electric field strength is 250 N/C. Substituting these values into the formula:

F = (3.50 × 10-6 C)(250 N/C)

F = 8.75 × 10-4 N.

The force is directed in the same direction as the electric field, which is due east. Therefore, the magnitude and direction of the force exerted on the charge is (8.75 × 10³ N) eastward.

User Bartosz Zasada
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