Calculate the force on charge at centre

Therefore, the only force acting on the charge will be due to the other charge placed at a distance of 2R from the center of the shell. In summary, the force exerted by the shell on the charge placed at the center of the shell will be zero, as there is no net electric field inside the conducting shell.
  • #1
gandharva_23
61
0
A charge Q is placed at the center of a spherical shell of radius R . Another charge Q is kept at a distance 2R from the centre of the shell . What will be the force exerted by shell on the charge placed at the centre of the shell ?

I know that the total force on the charge at centre will be force due to shell on on charge at centre (assuming the other charge was not there ) + the force charge at centre due to charge placed at 2R assuming the sphere was not there .(principal of superposition) . hence i can say that the total force on the charge at centre will be (K*q*q)/(2R)^2 . but i cannot understand how to calculate the force on charge at centre due to the shell alone when the other charge at 2R is also present . please help ...
 
Physics news on Phys.org
  • #2
gandharva_23 said:
A charge Q is placed at the center of a spherical shell of radius R . Another charge Q is kept at a distance 2R from the centre of the shell . What will be the force exerted by shell on the charge placed at the centre of the shell ?

I know that the total force on the charge at centre will be force due to shell on on charge at centre (assuming the other charge was not there ) + the force charge at centre due to charge placed at 2R assuming the sphere was not there .(principal of superposition) . hence i can say that the total force on the charge at centre will be (K*q*q)/(2R)^2 . but i cannot understand how to calculate the force on charge at centre due to the shell alone when the other charge at 2R is also present . please help ...
Assuming it is a conducting, uncharged shell, the charges will move around until they are all at equal potential. This means that the charges on the inside and outside surface will distribute so that there is always 0 field in the middle. Using Gauss' law, for each element of area of the shell, one can see that the net flux is 0. So I don't see how there would be a force by the shell on either charge.

AM
 

1. What is the formula for calculating the force on a charge at the centre?

The formula for calculating the force on a charge at the centre is F = k * (q1 * q2) / r^2, where F is the force, k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between the two charges.

2. How do I determine the direction of the force on a charge at the centre?

The direction of the force on a charge at the centre can be determined by using the right-hand rule. Point your thumb in the direction of the first charge, and your fingers in the direction of the second charge. Your palm will indicate the direction of the force.

3. Can the force on a charge at the centre be negative?

Yes, the force on a charge at the centre can be negative. This indicates an attractive force between two opposite charges.

4. What is the unit of measurement for the force on a charge at the centre?

The unit of measurement for the force on a charge at the centre is Newtons (N). This is the standard unit of force in the International System of Units (SI).

5. How does the distance between two charges affect the force on a charge at the centre?

The distance between two charges has an inverse-square relationship with the force on a charge at the centre. This means that as the distance increases, the force decreases, and vice versa.

Similar threads

  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
14
Views
10K
  • Introductory Physics Homework Help
Replies
4
Views
806
Replies
19
Views
858
  • Advanced Physics Homework Help
Replies
3
Views
500
  • Advanced Physics Homework Help
Replies
10
Views
3K
  • Advanced Physics Homework Help
Replies
4
Views
3K
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
1K
Replies
2
Views
1K
Back
Top