We need to find the force on a test charge Q at the center and force on Q in different situations
Answer:
(a)
We know, Electric force between two charges is directly proportional to the product of those two charges and inversely proportional to the square of the distance between two charges.
“F_e u221d \frac { qQ}{r^2}\\nF_e = \frac {1}{4\pi \epsilon _o} \frac { qQ}{r^2}”
Here r is the distance between the centre of the 12-sided polygon and its corner.
q and Q are the charges and “\epsilon_o” is a constant
If we observe the diagram, we can easily understood, at the corners we have the equal charge with the same distance from the centre of the polygon. So, both of these charges can apply the equal force on the test charge in the opposite directions.
So, sum of the total forces at test charge will be zero.
(b).
One of the charge in 12 q’s (means the one at “6 o’clock”) is removed, the force on the test charge at the centre is only one which is Opposite to “6 o’clock”, that is “12 o’clock”. We can write that force as
“F_e = \frac {1}{4\pi \epsilon _o} \frac { qQ}{r^2}”
(c). Now 13 equal charges, q, are placed at the corners of a regular 13–sided polygon.
Now, we can find the force on a test charge Q at the center
Here, also the force on the charge at the centre is zero. Because symmetry of the system.
d).
If one of the 13 q’s is removed from any corner of the Polygon, we can write a equation to net force due to the remaining 12 charges as
“F_e = \frac {1}{4\pi \epsilon _o} \frac { qQ}{r^2}”
