Answer :
Answer:
Electric force, F = 84.93 N
Explanation:
Given that,
Charge on particle 1, [tex]q_1=-7.97\ \mu C=-7.97\times 10^{-6}\ C[/tex]
Charge on particle 1, [tex]q_2=7.75\ \mu C=7.75\times 10^{-6}\ C[/tex]
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:
[tex]F=\dfrac{kq_1q_2}{d^2}[/tex]
k is the electrostatic constant
[tex]F=\dfrac{9\times 10^9\times 7.97\times 10^{-6}\times 7.75\times 10^{-6}}{(8.09\times 10^{-2})^2}[/tex]
F = 84.93 N
So, the magnitude of the force that one particle exerts on the other is 84.93 N.
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
[tex]F=\frac{Kqq'}{r^{2}}[/tex]
[tex]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}}[/tex]
F = 84.9 N
Thus, the force between the two charges is 84.9 N.