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4 years ago

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Two particles, one with charge − 7.97 μC −7.97 μC and one with charge 7.75 μC, 7.75 μC, are 8.09 cm 8.09 cm apart. What is the m

agnitude of the force that one particle exerts on the other?

2 answers:

mr Goodwill [35]4 years ago

4 0

Answer:

Electric force, F = 84.93 N

Explanation:

Given that,

Charge on particle 1, $q_1=-7.97\ \mu C=-7.97\times 10^{-6}\ C$

Charge on particle 1, $q_2=7.75\ \mu C=7.75\times 10^{-6}\ C$

The distance between charges, d = 8.09 cm

To find,

The magnitude of the force that one particle exerts on the other.

Solution,

The electric force of attraction or repulsion is given by the formula as:

$F=\dfrac{kq_1q_2}{d^2}$

k is the electrostatic constant

$F=\dfrac{9\times 10^9\times 7.97\times 10^{-6}\times 7.75\times 10^{-6}}{(8.09\times 10^{-2})^2}$

F = 84.93 N

So, the magnitude of the force that one particle exerts on the other is 84.93 N.

ZanzabumX [31]4 years ago

3 0

Answer:

84.9 N

Explanation:

q = 7.97 micro coulomb

q' = 7.75 micro coulomb

distance, d = 8.09 cm

The force between the two particles is given by

$F=\frac{Kqq'}{r^{2}}$

$F=\frac{9\times 10^{9}\times 7.97\times 10^{-6}\times 7.75 \times 10^{-6}}{\left (8.09\times 10^{-2} \right )^{2}}$

F = 84.9 N

Thus, the force between the two charges is 84.9 N.

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