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Alex17521 [72]
2 years ago
11

Two point charges, the first with a charge of 4.47 x 10-6 C and the second with a charge of 1.86 x 10-6 C, are separated by 17.4

mm. What is the magnitude of the electrostatic force experienced by charge 2
Physics
2 answers:
musickatia [10]2 years ago
5 0

Answer:

The force between the two charges is 247.15 N.

Explanation:

Charge, q = 4.47 x 10^-6 C

charge, q' = 1.86 x 10^-6 C

distance, d = 17.4 mm

Let the force is F.

The force is given by the Coulomb's law:

F = \frac{K q q'}{r^2}\\\\F =\frac{9\times 10^9\times 4.47\times 10^{-6}\times1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\F = 247.15 N

maria [59]2 years ago
3 0

Answer: 247.12\ N

Explanation:

Given

Magnitude of the charges

q_1=4.47\times 10^{-6}\ C

q_2=1.86\times 10^{-6}\ C

Distance between them d=17.4\ mm

As both charges are of same sign, they must repel each other

Force experienced by second charge is

\Rightarrow F_{21}=\dfrac{kq_1q_2}{d^2}\\\\\Rightarrow F_{21}=\dfrac{9\times 10^9\times 4.47\times 10^{-6}\times 1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\\Rightarrow F_{21}=\dfrac{74.82\times 10^{-3}}{302.76\times 10^{-6}}\\\\\Rightarrow F_{21}=0.2471\times 10^3\\\Rightarrow F_{21}=247.12\ N

Thus, charge 2 experience a force of 247.12\ N

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