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Chapter 18: Q27PE (page 664)

What is the magnitude and direction of an electric field that exerts a 2.00脳10-5Nupward force on a -1.75渭Ca charge?

Short Answer

Expert verified

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

Step by step solution

01

Given Data

Electric field Eexerts a force F=2.00脳10-5Non a charge =-1.75渭C.

02

Electric field

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

03

Magnitude and direction of electric field

From the definition of the electric field,

E=Fq

Here, F is the force F=2.00脳10-5N, and q is the test charge q=1.75渭C.

Substituting all known values,

E=2.00脳10-5N-1.75渭C=2.00脳10-5N-1.75渭C脳10-6C1渭C=11.43N/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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Earth has a net charge that produces an electric field of approximately\(150{\rm{ N/C}}\)downward at its surface. (a) What is the magnitude and sign of the excess charge, noting the electric field of a conducting sphere is equivalent to a point charge at its center? (b) What acceleration will the field produce on a free electron near Earth鈥檚 surface? (c) What mass object with a single extra electron will have its weight supported by this field?

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