3 years ago

Can someone please help me on this

Physics

1 answer:

Triss [41]3 years ago

4 0

Answer:

I am on the same question and I’m lost

Explanation:

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A test charge of +4 µC is placed halfway between a charge of +6 µC and another of +2 µC separated by 20 cm. (a) What is the magn

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(a) Magnitude: 14.4 N

(b) Away from the +6 µC charge

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As the test charge has the same sign, the force that the other charges exert on it will be a repulsive force. The magnitude of each of the forces will be:

$F_e = K\frac{qq_{test}}{r^2}$

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q and qtest is the charge of the particles, and r is the distance between the particles.

Let's say that a force that goes toward the +6 µC charge is positive, then:

$F_e_1 = K\frac{q_1q_{test}}{r^2}=-9*10^9 \frac{Nm^2}{C^2} \frac{6*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =-21.6 N$

$F_e_2 = K\frac{q_2q_{test}}{r^2}=9*10^9 \frac{Nm^2}{C^2} \frac{2*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =7.2 N$

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