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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.

Respuesta :

Answer:

-2370000 N force acts on the charge particle    

Explanation:

We have given electric field E = 790000 N/C

Charge [tex]q=-3\mu C=-3\times 10^{-6}C[/tex]

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

[tex]F=qE[/tex], here q ia charge and E is electric field

So force [tex]F=-3\times 10^{-6}\times 790000=-2370000N[/tex]

So -2370000 N force acts on the charge particle  

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