Question

A positively charged oil drop remains stationary in an electric field between two horizontal plates separated by distance of 1 cm for the charge of the job is 9.6 into 10 raise to minus 6 coulomb and mass of that refers to raise to minus 7 gram what is the potential difference between plates

Answer 1

here, electrostatic energy consumed to balance weight of oil drop.

e.g., workdone due to weight = workdone due to potential difference V

or, (mg) × x = qV

given, m = 10^-7 g = 10^-10 kg , g = 9.8 m/s² , q = 9.6 × 10^-6 C , x = 1cm =0.01m

now, 10^-7 kg × 9.8 m/s² × 0.01 m= 9.6 × 10^-6 C × V

or, V = 9.8 × 10^-9/(9.6 × 10^-6)

V = 1.02 × 10^-3 volts

hence, potential difference between plates is 1.02 × 10^-3 volts.

Answer 2

This is your answer

thank you