E@(yen)?$(pounds)a positive test charge of magnitude 2.40X10U-8 C experiences a force of 1.50X10U-3 N toward the east. What is the electric field at the position of the
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Answer:
The intensity of an electric field is equal to the force divided by the charge. So the field strength equals 40 N/C in this case.
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Other answers
The intensity of an electric field is equal to the force divided by the charge. So the field strength equals 40 N/C in this case.
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