Each of these charged spheres exert a force on each that is equal and opposite on one another in the x-direction.  The tension force on each of these charged spheres is equal to the force from the other sphere in the x-direction and the force of gravity in the y-direction.  We will be able to use this information to determine the charge on the electroscope.

First we must analyze the forces on the charges and determine the force on the charges.  We know that

\(\displaystyle tan\theta = {F_{sphere}}{F_{gravity}}\)

Rearranging this we get

\(\displaystyle F_{gravity}tan\theta = F_{sphere}\)

\(\displaystyle (0.020kg)(9.8m/s^2)tan(25) = F_{sphere}\)

\(\displaystyle F_{sphere} = 0.091N\)

Next we can use Coulomb’s law to determine the magnitude of each charge.

\(\displaystyle F_{sphere} = \frac{kq_1q_2}{d^2}\)

In order to figure out the distance between the two spheres we can trigonometry with the length of the string.

\(\displaystyle sin\theta = \frac{opp}{hyp}\)

\(\displaystyle (0.8m)(sin(25) = opp\)

\(\displaystyle opp = 0.34m\)

Therefore the distance between the two charges is double this value.

\(\displaystyle d = 0.68m\)

\(\displaystyle 0.091N = \frac{(9x10^9)(\frac{Q}{2})(\frac{Q}{2})}{(0.68m)^2}\)

\(\displaystyle 0.091N = \frac{(2.25x10^)Q^2}{0.4624}\)

\(\displaystyle (0.091N)(0.4624) = (2.25x10^9)Q^2\)

\(\displaystyle 0.042 = (2.25x10^9)Q^2\)

\(\displaystyle 1.87x10^11 = Q^2\)

\(\displaystyle Q = 4.32x10^{-6}C\)

First let us determine the force of the two charges on each other.

\(\displaystyle F = \frac{kq_1q_2}{d^2}\)

\(\displaystyle F = \frac{k(-Q)(-3Q)}{l^2}\)

\(\displaystyle F = \frac{k3Q^2}{l^2}\)

The easiest way to analyze this is to assume all of these are in a straight line and we have a charged placed somewhere along the same x-axis between the charges that keeps them all from moving.  Let us assume the smaller charge is at the origin and the distance to the fixed charge is some distance \(\displaystyle r\).

\(\displaystyle F = \frac{k(-Q)(xQ)}{r^2}\)

Let us also assume that the larger charge is a distance l away from the origin (away from the smaller charge) and therefore a distanc l-r away from the fixed charge.

\(\displaystyle F = \frac{k(-3Q)(xQ)}{(l-r)^2}\)

To be in equilibrium we know that the net force on the smaller charge must equal \(\displaystyle 0\).

\(\displaystyle \frac{k3Q^2}{l^2} + \frac{k(-Q)(xQ)}{(r)^2} = 0\)

We also know that the net force on the larger charge must equal \(\displaystyle 0\)

\(\displaystyle \frac{k3Q^2}{l^2} + \frac{k(-3Q)(xQ)}{(l-r)^2} = 0\)

We can set these two equations equal to each other.

\(\displaystyle \frac{k3Q^2}{l^2} + \frac{k(-Q)(xQ)}{(r)^2} = \frac{k3Q^2}{l^2} + \frac{k(-3Q)(xQ)}{(l-r)^2}\)

The force between the charges can be cancelled since it is on both sides.

\(\displaystyle \frac{k(-Q)(xQ)}{(r)^2} = \frac{k(-3Q)(xQ)}{(l-r)^2}\)

The k and Q values can also be cancelled out on both sides.

\(\displaystyle \frac{-1}{(r)^2} = \frac{-3(l-r)^2} {[?]}\)

Cross multiply on both sides

\(\displaystyle (l-r)^2 = 3r^2\)

\(\displaystyle l^2 - 2lr +r^2 = 3r^2\)

\(\displaystyle l^2 - 2lr -2r^2 = 0\)

This is a quadratic that needs to be solved with a quadratic formula.

\(\displaystyle r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

\(\displaystyle r =\frac{-(-2l)\pm\sqrt{(2l)^2-4(-2)(l^2)}}{2(-2)}\)

\(\displaystyle r = \frac{2l\pm\sqrt{4l^2+8l^2}}{-4}\)

\(\displaystyle r = \frac{2l\pm\sqrt{12l^2}}{-4}\)

\(\displaystyle r = \frac{2l\pm2\sqrt{3}l}{-4}\)

\(\displaystyle r = \frac{l\pm\sqrt{3}l}{-2}\)

\(\displaystyle r = \frac{-l\pm\sqrt{3}l}{2}\)

So the two options for the \(\displaystyle r\) value would be

\(\displaystyle r = \frac{-2.73l}{2} = -1.37l\)

\(\displaystyle r = \frac{0.73l}{2} = 0.366l\)

Of these two, the one that makes the most sense based on our assumptions is the \(\displaystyle 0.366l\) as this would be in the middle of two charges.

We can now go back to our first equation with the smaller charges

\(\displaystyle \frac{k3Q^2}{l^2} + \frac{k(-Q)(xQ)}{(r)^2} = 0\)

We can plug in our value for the \(\displaystyle r\).

\(\displaystyle \frac{k3Q^2}{l^2} +\frac{k(-Q)(xQ)}{(0.366l)^2} = 0\)

The \(\displaystyle k, Q^2\) and \(\displaystyle l^2\) can be cancelled out

\(\displaystyle 3 + \frac{x}{(0.366)^2} = 0\)

\(\displaystyle 3 +\frac{x}{.134} = 0\)

\(\displaystyle -3 = \frac{x}{.134}\)

\(\displaystyle -0.4 = x\)

The charge must be placed \(\displaystyle 0.366l\) away from the smaller charge with a magnitude of \(\displaystyle -0.4Q\).

We can calculate the force from each of the two charges on the center charge using Coulomb’s Law

\(\displaystyle F = \frac{kq_1q_2}{d^2}\)

The first charge on the middle charge

\(\displaystyle F = \frac{(9x10^9)(70x10^{-6})(45x10^{-6})}{(0.4m)^2}\)

\(\displaystyle F = 177N\)

The last charge on the middle charge

\(\displaystyle F = \frac{(9x10^9)(45x10^{-6})(-100x10^{-6})}{(0.4m)^2}\)

\(\displaystyle F = -253N\)

To find the total force we need to add both of these forces together.

\(\displaystyle F = 177N + (-253N)\)

\(\displaystyle F = -76N\)

Since the charges are opposite, the center charge +Q is going to be attracted to each of the charges in the corner of the square.  Since each charge is equidistant from the center charge, they will each exert 4N of force on the charge in the center.  However, since each corner charge is pulling the center charge equally and oppositely, the net force on the system is equal to 0 and the charge will not move.  All of the forces cancel out with one another.

In Coulomb’s Law, an increase in the distance will cause a decrease in the magnitude of the electrical force between them.  Since this is an example of an inverse square law, doubling the distance will reduce the force by a factor of 4.

Electric forces can be attractive or repulsive because charges may be positive or negative. In the case for gravitational forces, there are only attractive forces because mass is always positive.

We can calculate this using Coulomb’s Law

\(\displaystyle F = \frac{kq_1q_2}{d^2}\)

\(\displaystyle F = \frac{(9x10^9)(25x10^{-6})(25x10^{-3})}{(0.16m)^2}\)

\(\displaystyle F = 219727N = 220,000N\)

In Coulomb's law, an increase in both interacting charges will cause an increase in the magnitude of the electrical force between them. Specifically if the magnitude of both interacting charges is doubled, this will quadruple the electrical force.

We will need to use Coulomb’s Law to analyze the force on the 3Q charge from all the other forces.  We will then summarize the net force in the \(\displaystyle x\) and \(\displaystyle y\) direction to determine the force on the 3Q charge.

In the x-direction

Force from 4Q on 3Q

\(\displaystyle F = \frac{k4Q3Q}{l^2}\)

Force from Q on 3Q in the x direction

\(\displaystyle F =\frac{kQ3Q}{(\sqrt{2}l)^2}\)

\(\displaystyle F = \frac{k3Q^2}{2l^2}\)

Add these together in the x-direction 

\(\displaystyle F = \frac{k12Q^2}{l^2} + \frac{k3Q^2}{2l^2}\)

\(\displaystyle F = \frac{27kQ^2}{2l^2}\)

In the y-direction

From from 2Q on 3Q

\(\displaystyle F = \frac{k2Q3Q}{l^2}\)

\(\displaystyle F = \frac{k6Q^2}{l^2}\)

Force from Q on 3Q in the y direction

\(\displaystyle F =\frac{kQ3Q}{(\sqrt{2}l)^2}\)

\(\displaystyle F = \frac{k3Q^2}{2l^2}\)

Add these together in the y-direction 

\(\displaystyle F = \frac{k6Q^2}{l^2} + \frac{k3Q^2}{2l^2}\)

\(\displaystyle F = \frac{15kQ^2}{2l^2}\)

Now we can find the resultant of these sides using the Pythagorean theorem.

\(\displaystyle \frac{27kQ^2}{2l^2}^2 + \frac{15kQ^2}{2l^2}^2 = c^2\)

\(\displaystyle \frac{729k^2Q^4} {4l^4} + \frac{225k^2Q^4}{4l^2} = c^2\)

\(\displaystyle \frac{954k^2Q^4}{4l^4} = c^2\)

\(\displaystyle \sqrt{\frac{954k^2Q^4}{4l^4}} = \sqrt{c^2}\)

\(\displaystyle \frac{30.9kQ^2}{2l^2} = c\)

\(\displaystyle 15.5\frac{kQ^2}{l^2} = c\)

The power in a circuit is equal to the current times the voltage.

\(\displaystyle P = IV\)

\(\displaystyle P = (240x10^{-3}A)(3v)\)

\(\displaystyle P = 0.72W\)