Question

3. Let's see how the strength of the electrical force compares to the gravitational force. If two 10 g ice cubes of pure water were a distance r apart. If we placed the same amount of excess charge on each ice cube, how much excess charge would we need so that the magnitude of the electrical force was the same as the gravitational force between them?

Answer 1

3.

Gravitational force is given by:

F1 = G*m1*m2/R^2

G = Gravitational constant = 6.67*10^-11

m1 = m2 = mass of ice cubes = 10 gm = 10*10^-3 kg

R = distance between both mass = r

So,

F1 = 6.67*10^-11*(10*10^-3)^2/r^2

F1 = 6.67*10^-15/r^2 N

Now electrostatic force is given by:

F2 = k*q1*q2/R^2

k = electrostatic constant = 9*10^9

q1 = q2 = Q = ?

R = r

So,

F2 = 9*10^9*Q^2/r^2

Now given that we need amount of charge on ice cube when

Gravitational Force = Electrostatic Force

F1 = F2

6.67*10^-15/r^2 = 9*10^9*Q^2/r^2

Q^2 = (6.67*10^-15)/(9*10^9)

Q = sqrt [(6.67*10^-15)/(9*10^9)]

Q = 8.6*10^-13 C = 0.86*10^-12 C

(1*10^-12 C = 1 pC)

Using above

Q = Charge on each ice cube = 0.86 pC

Since gravitation force is always attractive, So electrostatic force should be repulsive. And Since we are placing excess charge on each ice cube so charge will be negative.

Q = Charge on each ice cube = -0.86 pC

Total excess charge required = 2*Q = 2*0.86 pC = 1.72 pC

Please Upvote.