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Suppose you want to determine the electric field in a certain region of space. You have a small object of known charge and an instrument that measures the magnitude and direction of the force exerted on the object by the electric field. How would you determine the magnitude and direction of the electric field if the object were (a) positively charged and (b) negatively charged? (a) The object has a charge of +20.0μC and the instrument indicates that the electric force exerted on it is
40.0μN due east. What are the magnitude and direction of the electric field? (b) What are the magnitude and direction of the electric field if the object has a charge of −10.0μC and the instrument indicates that the force is 20.0μN due west?

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Answer:

2 N/C direction of the force

-2N/C opposite to the direction of the force

Explanation:

E = Electric field

q = Charge

Electrical force is given by

[tex]F=Eq\\\Rightarrow E=\dfrac{F}{q}\\\Rightarrow E=\dfrac{40}{20}\\\Rightarrow E=2\ N/C[/tex]

The magnitude of the force is 2 N/C

The force acting on the charge is positive so the direction of the electric field is positive.

[tex]F=Eq\\\Rightarrow E=\dfrac{F}{q}\\\Rightarrow E=\dfrac{20}{-10}\\\Rightarrow E=-2\ N/C[/tex]

The magnitude of the force is 2 N/C

The direction of the electric field is negative and opposite to the direction of the force as the charge is negative.

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