Question

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.

Answer 1

Answer:

-2370000 N force acts on the charge particle    

Explanation:

We have given electric field E = 790000 N/C

Charge $q=-3\mu C=-3\times 10^{-6}C$

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 $F=-3\times 10^{-6}\times 790000=-2370000N$

So -2370000 N force acts on the charge particle