a 7.50-nc point charge is located 1.80 m from a 4.20-nc point charge. (a) find the magnitude of the electric force that one particle exerts on the other. (b) is the force attractive or repulsive?

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24-10-2022
Physics

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a 7.50-nc point charge is located 1.80 m from a 4.20-nc point charge. (a) find the magnitude of the electric force that one particle exerts on the other. (b) is the force attractive or repulsive?

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bharathparasad577 bharathparasad577 · Answer 1 · 2022-10-28T03:46:18-04:00

The electric force is 8.370 x 10^9N.

We need to know about the electric force to solve this problem. The electric force produced by charges interacting each other, whose strength is given by the equation

F = k.Q1 . Q2/r²

where F is the electric force, Q1 and Q2 is the charge and r is the radius.

From the question above, we know that

k = 9 x 10⁹ Nm²/C²

Q1 =e = 7.50 x 10^-9 C

Q2 = e = 4.02 x 10^-9 C

r = 1.80 m

The electric force will attract each other when the charges are different and will repel each other when the charges are same.

F = k.Q1 . Q2/r²

F = 9 x 10⁹ . 7.50 x 10^-9 . 4.02 x 10^-9/(1.80)²

F = 8.37 x 10^9 N

Find more on electric force at: brainly.com/question/17692887

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