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Kay [80]
3 years ago
9

Two charged particles are a distance of 1.62 m from each other. One of the particles has a charge of 7.10 nc, and the other has

a charge of 4.42 nc. (a) What is the magnitude (in N) of the electric force that one particle exerts on the other? (b) is the force attractive or repulsive? O attractive O repulsive
Physics
1 answer:
k0ka [10]3 years ago
8 0

Answer:

A. F=107.6nN

B. Repulsive

Explanation:

According to coulombs law, the force between two charges is express as

F=(Kq1q2) /r^2

If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.

Note the constant K has a value 9*10^9

Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m

If we substitute values we have

F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)

F=(282.4×10^-9)/2.6244

F=107.6×10^-9N

F=107.6nN

B. Since the charges are both positive, the force is repulsive

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\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

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\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

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