A particle with a charge of -3.77 x 106 C is separated from a particle of charge

Question

A particle with a charge of -3.77 x 106 C is separated from a particle of charge 1.87 x106 C by a distance of 0.0359 meters. What is the magnitude and direction of the electrical force that the 1.87 x106 C charge exerts on the 3.77 x106 C charge? Electrical force on negative charge Number O away from the positive charge NO toward the positive charge What is the magnitude and direction of the electric field at the 1.87 x106 C charge's location due to the -3.77 x106 C charge? Electrical field at positive charge's location Number O toward the negative charge away from the negative charge N/C

Explanation / Answer

Expression for the electric force between the two charges is given by -

F = (k*q1*q2) / r^2

Here, k = constant = 9 x 10^9 N*m^2 / C^2

q1 = 1.87 x 10^-6 C

q2 = -3.77 x 10^-6 C

r = 0.0359 m

Substitute the given values in the above expression -

F = (9 x 10^9 x 1.87 x 10^-6 x (-3.77 x 10^-6)) / 0.0359^2 = -49.23 N

So, magnitude of the electric force = 49.23 N

And its direction is towards the positive charge.

Second Part -

Expression force the electric field due to a charge is given by -

E = k*q / r^2

Given that -

k = constant = 9 x 10^9 N*m*2/C^2

q = -3.77 x 10^-6 C

r = 0.0359 m

So, E = (9 x 10^9 x (-3.77 x 10^-6)) / 0.0359^2 = - 2.633 x 10^7 N/C

So, magnitude of the electric field = 2.633 x 10^7 N/C

And, its direction is toward the negative charge.

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