Question

A 2.0 g metal cube and a 4.0 g metal cube are 6.0 cm apart, measured between their centers, on a horizontal surface. For both, the coefficient of static friction is 0.65. Both cubes, initially neutral, are charged at a rate of 7.0 nC/s.
a. How long after charging begins does one cube begin to slide away?
b. Which cube moves first?

Answer 1

Answer:

a) t=10.2s

b) The 2g-cube moves first

Explanation:

Since the electric force is the same on both cubes and so is the coefficient of static friction, the first one to move will be the one with less mass.

So, on the 2g-cube the sum of forces are:

$\left \{ {{Ff-Fe=0} \atop {N-m*g=0}} \right.$

Replacing the friction on the first equation:

$\mu*m*g-Fe=0$  Thus   $Fe=\mu*m*g=12.74*10^{-3}N$

The electric force is:

$Fe = \frac{K*q^2}{d^2}$  Solving for q:

q=71.44nC

This amount divided by the rate at which they are being charged:

t = 71.44nC / 7nC/s = 10.2s