Question

A 7.94 nC charge is located 1.79 m from a 4.06 nC point charge. (a) Find the magnitude of the electrostatic force that one charge exerts on the other. N (b) Is the force attractive or repulsive? attractive repulsive

Answer 1

Answer:

$9.044824943\times 10^{-8}\ N$

repulsive

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

$q_1$ = 7.94 nC

$q_2$ = 4.06 nC

r = Distance between the particles = 1.79 m

Electrical force force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 7.94\times 10^{-9}\times 4.06\times 10^{-9}}{1.79^2}\\\Rightarrow F=9.044824943\times 10^{-8}\ N$

The magnitude of the electrostatic force is $9.044824943\times 10^{-8}\ N$

Both the paricles have positive charges this means they will repel each other.