Posted by Abdullahi on .
Two insulated charged metallic spheres A & B have their centres separated by a distance of 50cm in air. Calculate the force of electrostatic repulsion between them. If the charge on each is +5.0*10^-8c. The radii of A & B are negligible compared to their distance of separation.
( b). What is the force of repulsion if (i)the charge doubled and between them is halved.
(ii) the two spheres are placed in water of dielectric constant 81? [(a)F=9*10^-5N,(b)(i)F=1.4*10^-3N, (ii)F=1.1*10^-6N]
a) use coulomb's law and your calculator.
b) Double charges? force goes up by 2*2. Distance is halved? Force goes up by 4
total force increase: 4*4
c) dielectric constant will reduce the force by a factor of 1/81
I didn't check your answers.