Question

Two spherical drops of mercury each have a charge of 9.0×10⁻² nC and a potential of 310 V at the surface. The two drops merge to form a single drop.

What is the potential at the surface of the new drop?

Express your answer to two significant figures and include the appropriate units

Answer 1

Electric potential due to point charge is given by

V=KQ/r

Therefore

=>r=KQ/V =(9*10⁹)(9*10^-2*10^-9)/310

r=2.613*10^-3 m

After two drops merge into single drop ,

V' =2V

(4pi/3)r'²=2(4pi/3)r³

r' =(2)^1/3(2.613*10^-3)=3.292*10^-3 m

Total charge

Q'=2Q =2*(9*10^-11)=1.8*10^-10 C

Therefore potential at the surface of the new drop is

V'=KQ'/r'=(9*10⁹)(1.8*10^-10)/(3.292*10^-3)

V' = 492.1 Volts =4.9*10² Volts (approx)