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1- A positive charge of 3x10-7 is located in a field of 27N/C directed toward the south. What is the force acting on the charge?



2- A positive test charge of 5x10-6Cis in an electric field that exerts a force of 2x10-4N on it. What is the magnitude of the electric field at the location of the test charge?

Respuesta :

Answer:

1)  8.1x[tex]10^{-6}[/tex] N

2) 4.0x[tex]10^{1}[/tex] N/C

Explanation:

1) E = F/q ---> F = Eq = (27 N/C) x (3x[tex]10^{-7}[/tex])

= 8.1x[tex]10^{-6}[/tex] N

2) E = [tex]\frac{F}{q}[/tex] = [tex]\frac{2.0X10^{-4} }{5.0X10^{-6} }[/tex]

= 4.0x[tex]10^{1}[/tex] N/C

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