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An electric field of 790,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -3.00 uC at this spot? (14C = 10 6C) Give your answer in Sl unit rounded to two decimal places.

User Leen
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1 Answer

5 votes

Answer:

-2370000 N force acts on the charge particle

Step-by-step explanation:

We have given electric field E = 790000 N/C

Charge
q=-3\mu C=-3* 10^(-6)C

We know that force on any charge particle due to electric field is given by


F=qE, here q ia charge and E is electric field

So force
F=-3* 10^(-6)* 790000=-2370000N

So -2370000 N force acts on the charge particle

User Dave Greilach
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