The electrostatic force on a small sphere of charge 0.4 μC due toanother small sphere of charge –0.8 μC in air is 0.2 N. (a) What isthe distance between the two spheres? (b) What is the force on thesecond sphere due to the first?
Given, the charge on the sphere is 0.4 μC , the charge on the other sphere is − 0.8 μC and the electrostatic force due to both spheres is 0.2 N .
(a)
The electrostatic force between the charges is given by the equation,
F = 1 4 π ε 0 × q 1 q 2 r 2 r 2 = 1 4 π ε 0 × q 1 q 2 F
Substituting the values in the above equation, we get:
r 2 = 9 × 10 9 N ⋅ m 2 / C − 2 × 0.4 × 10 − 6 C × 0.8 × 10 − 6 C 0.2 N r = 0.12 m
Hence, the distance between the two spheres is 0.12 m.
(b)
The electrostatic force will be the same on both the charges. Hence, the force of attraction between the spheres is the same and is equal to 0.2 N .
Given, the charge on the sphere is
(a)
The electrostatic force between the charges is given by the equation,
Substituting the values in the above equation, we get:
Hence, the distance between the two spheres is 0.12 m.
(b)
The electrostatic force will be the same on both the charges. Hence, the force of attraction between the spheres is the same and is equal to
