Aspherical drop of water carrying a charge of 38 pc has a potential of 710 v at its surface (with v = 0 at infinity). (a) what is the radius of the drop? (b) if two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop?

a) r= 4.82 * 10^-4 m ; b) 1420 V

Explanation: In order to solve this problem we have to take into account that potential for a sphere respec to V=0 at the infinity, which is given by:

V=k*Q/r where r is the radius of the drop

then we have

r=k*Q/V=9*10^9*38pC/710V= 4.82 * 10^-4 m

Finally if we join two drop to form one with the same radius but with twice charge the resultant potential is:

V= k*2*Q/(r)= 710*2= 1420 V

a coal burning power plant

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min