Question

A 3.0 mg bead with a charge of 2.9 nC rests on a table. A second bead, with a charge of -5.3 nC is directly above the first bead and is slowly lowered toward it. What is the closest the centers of the two beads can be brought together before the lower bead is lifted off the table?

Answer 1

6.86 cm

Step-by-step explanation:

Given:

• q = charge on the first bead on the table=
  `2.9\ nC = 2.9* 10^(-9)\ C`
• m = mass of bead on the table =
  `3.0\ mg = 3.0* 10^(-6)\ kg`
• Q = charge on the second bead =
  `-5.3\ nC = -5.3*10^(-9)\ C`

Assume:

• r = the closest distance between the centers of the beads
• F = electrostatic force of attraction between the two beads
• W = weight of the first bead
• g = acceleration due to gravity = 9.8\ m/s^2
• N = normal force on the first bead

When the first bead rests on the table, then electrostatic force due to the second bead acts on it in the upward direction, Normal force acts in the upward direction and its weight in the downward direction.

So, using Newton's second law on the first bead resting on the table, we have

`F+N-W=0\\`

At the closest distance of the second bead to the first bead, it just lifts off the table and the normal force becomes zero.

`\therefore F-W=0\\\Rightarrow F=W\\\Rightarrow (kqQ)/(r^2)=mg\\\Rightarrow r^2=(kqQ)/(mg)\\\Rightarrow r^2=(9* 10^9* 2.9* 10^(-9)* 5.3* 10^(-9))/(3* 10^(-6)* 9.8)\\\Rightarrow r^2=4.70* 10^(-3)\\\textrm{Taking square root on both the sides}\\r = \pm 0.0686\ m\\\textrm{Since the distance is never negative}\\\therefore r = 0.0686\ m\\\Rightarrow r = 6.86\ cm`

Hence, the centers of the two beads must be brought closest to 6.86 cm before the lower bead is lifted off the table.