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College Physics (Urone)
Found in: Page 664

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Short Answer

What is the magnitude and direction of an electric field that exerts a 2.00×10-5 N upward force on a -1.75 μCa charge?

The magnitude of the electric field is 11.43 N/C , and the direction is downward.

See the step by step solution

Step by Step Solution

Step 1: Given Data

Electric field E exerts a force F=2.00×10-5Non a charge =-1.75 μC.

Step 2: Electric field

The force per unit positive test charge is known as the intensity of the electric field.

Step 3: Magnitude and direction of electric field

From the definition of the electric field,


Here, F is the force F=2.00×10-5N , and q is the test charge q=1.75 μC .

Substituting all known values,

E=2.00×10-5N-1.75 μC =2.00×10-5N-1.75 μC×10-6C1 μC =11.43 N/C

Hence, the magnitude of the electric field is 11.43 N/C.

The negative sign of the electric field indicates that the direction of electric field is in the opposite direction of force. Since, the force is directed upward. Hence, the electric field is directed downward.


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