# (1) Two particles, each of mass m and carrying charge Q, are separated by some distance.If they are in equilibrium under mutual gravitational and electrostatic forces,then Q/m (in C/Kg) is of the order of(a) 10-5(b) 10-10(c) 10-15(c) 10-20

Mujahid Ahmed
42 Points
10 years ago

K Q2/r2 = Gm2/r2

Q2/m2 = G/K

= 6,6 x 10-11/9 x 109

= 10-20

Approve if i helped :)

Abhishekh kumar sharma
34 Points
10 years ago

10^-15

anurag hundal
21 Points
10 years ago

guess its c!

Jatin Singh
24 Points
7 years ago
repulsive electrostatic force:                         attractive gravitational force:
F1=kQ^2/r^2                                                   F2=Gm^2/r^2

for equilibrium :F1=F2
kQ2/r2=Gm2/r2
=> Q/m = (6.6*10^-11/9*10^9)1/2
=> Q/m=0.8*10-10
Hence, ORDER OF Q/m=10-10

askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem below.
repulsive electrostatic force(since charges are same in magnitude and sign):
F1 = kQ2/r2
attractive gravitational force:
F2 = Gm2/r2
for equilibrium :F1 = F2
kQ2/r2 = Gm2/r2
=> Q/m = (6.6*10-11/9*109)1/2
=> Q/m=0.8*10-10
Hence, ORDER OF Q/m=10-10
Hence (b) is the correct option.

Hope it helps.
Thanks and regards,
Kushagra