# Electrostatics problems

Hello,

The problem states:

Three small spheres with charge 2.00 mC are arranged in a line, with sphere 2 in the middle. Adjacent spheres are initially 8.0 cm apart. The spheres have masses m_1 = 20.0 g, m_2 = 85.0 g, and m_3 = 20.0 g. Their radii are much smaller than their seperation. The three spheres are released from rest.

It asks to:

1. Find the acceleration of sphere 1 just after it is released?

2. What is the speed of sphere 1 when the spheres are far apart?

3. What is the speed of sphere 2 when the speres are far apart?

4. What is the speed of sphere 3 when the spheres are far apart?

I understand that it is again a conservation of energy problem. I can calculate initial potential energy based on:

U = qV = qk(q_1/r_1 + q_2/r_1 + q_3/r_1)

I tried finding the electric field using this potential because of:

V = E*d with d = 0.08 m

Then I tried to find the Electrical force according to: F = q*E

Then I tried finding the acceleratoin given by: a = F/mass

I got 2.80 * 10^8 m/s^2 which is not the right answer.

Thank you!!!

Â© BrainMass Inc. brainmass.com March 4, 2021, 7:22 pm ad1c9bdddfhttps://brainmass.com/physics/acceleration/electrostatics-problems-97113

#### Solution Preview

You can indeed solve the problem using conservation of energy. But you need to use the correct expression for the potential energy. If you have three particles with charges q_1, q_2 and q_3 at positions r_1, r_2 and r_3 respectively, then the potential energy is:

U(r_1, r_2, r_3) = k q_1 q_2/|r_1 - r_2| + k q_1 q_3/|r_1 - r_3| + k q_2 q_3/|r_2 - r_3|

Here |x| denotes the absolute value of x. Note that this formula is also valid in more than one dimension. Then the r_j are vectors and |x| denotes the length of the vector, so the |r_i - r_j| would then again be the distance from i to j.

The force that a particle experiences is minus the derivative w.r.t. its position coordinate. So, if you differentiate U w.r.t. r_1, you get minus the force exerted in the positive r_1 direction. To calculate the derivatives it is convenient to rewrite 1/|x| as sign(x)/x. Sign(x) = 1 if x> 0 and -1 if x < 0. If you calculate the derivative you find for the force exerted on charge 1 in the positive

F_1 = k q_1 q_2 sign(r_1 - r_2)/(r_1 - r_2)^2 + k q_1 q_3 sign(r1 - r3)/(r_1 - ...

#### Solution Summary

Detailed solutions are given.