electrostatics – Worksheet 1 problem solution (Q2)
Question 2) In the picture below, X is a small negatively charged sphere with a mass of 10kg. It is suspended from the roof by an insulating rope which makes an angle of 60o with the roof. Y is a small positively charged sphere that has the same magnitude of charge as X. Y is fixed to the wall by means of an insulating bracket.
Assuming the system is in equilibrium, what is the magnitude of the charge on X?
![](https://physicsteacher.in/wp-content/uploads/2022/03/image-82.png)
[ here is the complete worksheet: electrostatics worksheet 1 ]
Solution:
determining the charge on X – to determine their charges it’s required to know the force between X and Y & we can use Coulomb’s Law as we know the distance between them.
So, firstly, we need to determine the magnitude of the electrostatic force between X and Y.
The distance between X and Y is 50cm = 0.5m, and the mass of X is 10kg.
Draw the forces on X (with directions) and label.
![](https://physicsteacher.in/wp-content/uploads/2022/03/image-88.png)
Determine the magnitude of the electrostatic force (FE). Since nothing is moving (the system is in equilibrium) the vertical and horizontal components of the forces must cancel. Thus:
![](https://physicsteacher.in/wp-content/uploads/2022/03/image-89.png)
The only force we know is the gravitational force Fg = mg. Now we can calculate the magnitude of T from above:
![](https://physicsteacher.in/wp-content/uploads/2022/03/image-90.png)
This means that FE is:
![](https://physicsteacher.in/wp-content/uploads/2022/03/image-91.png)
Now that we know the magnitude of the electrostatic force between X and Y, we can calculate their charges using Coulomb’s Law. Don’t forget that the magnitudes of the charges on X and Y are the same: |QX| = |QY|. The magnitude of the electrostatic force is:
![](https://physicsteacher.in/wp-content/uploads/2022/03/image-92.png)
Thus the charge on X is -1.27 x 10-4 C
[ here is the complete worksheet: electrostatics worksheet 1 ]