Force on a point charge due to a sphere

In summary, the problem involves an insulated conducting sphere with a radius of R and a total charge of Q, in the field of a point charge q of the same sign. The objective is to calculate and plot the force exerted by the sphere on q as a function of distance from the center, with a particular case of q/Q = 0.1 where the direction of force reverses. The solution involves using the legendre solutions to laplace's equation and applying boundary conditions to find the potential and electric field of the sphere. An alternative method using the image method is also suggested.
  • #1
andre220
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1

Homework Statement


An insulated conducting sphere of radius ##R##, carrying a total charge of ##Q##, is in the field of a point charge ##q## of the same sign. Assume ##q\ll Q##. Calculate and plot the force exerted by the sphere on ##q## as a function of distance from the center. In particular, for ##q/Q = 0.1##, calculate the point at which the direction of force reverses.

Homework Equations


The electric field due a point charge $$E_\mathrm{pt} = \frac{q}{4\pi\epsilon_0 r^2}$$
$$\Phi(r,\theta) = \sum\limits_{l=0}^\infty \left(A_l r^l + \frac{B_l}{r^{l+1}}\right) P_l\left(\cos{\theta}\right)$$
$$\vec{E} = -\vec{\nabla}\Phi$$

The Attempt at a Solution


So since the solution is wanted as a function of distance from the center of the sphere, I am using the legendre solutions to laplace's equation. The boundary conditions are ##\Phi(R,\theta) = 0## and ##E_\mathrm{in} -E_\mathrm{out} = -\sigma/\epsilon_0 \implies E_\mathrm{point} - \nabla_r \Phi =-\frac{4\pi R^2}{\epsilon_0}##
Then once the boundary conditions are satisfied then the force is just ##F = qE_\mathrm{sphere}##.

So for the potential of the sphere:
$$\Phi(r) = A_0 + \frac{B_0}{r}$$
$$\Phi(R) = 0 = A_0 + \frac{B_0}{R}\implies B_0 = -A_0 R\implies \Phi(r) = A_0\left(1-\frac{R}{r}\right)$$
$$E_\mathrm{sphere} = -\frac{2 A_0 R}{r^2}$$
Then, applying the boundary conditions at the surface of the sphere:
$$\frac{q}{4\pi\epsilon_0 R^2} + \frac{2 A_0}{R} = -\frac{4\pi R^2 Q}{\epsilon_0}\implies A_0 = \frac{-q + 16\pi^2 R^4 Q}{8\pi\epsilon_0 R}$$

Then,
$$E_\mathrm{sphere} = \frac{1}{r^2}\left(\frac{q + 16\pi^2 Q R^4}{4\pi\epsilon_0}\right)$$
Then from here I would just do force by multiplying by ##q## of the point charge and then find the point at which the force reverses.

Alot to take in, but does this seem correct thus far?
 
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  • #2
andre220 said:
So for the potential of the sphere:
$$\Phi(r) = A_0 + \frac{B_0}{r}$$
Where does that come from? I think the idea of the conducting sphere is the non-uniform charge distribution. The electric field and therefore the potential should depend on θ.
 
  • #3
Right, then I don't see how I can get the constants ##A_1, B_1## with only two boundary conditions.
Thats why I only chose to keep ##l = 0## terms.
 
  • #4
There are formulas how a charge q at some distance L will influence charges on a conducting sphere. It's possible to derive them but I guess that is not necessary here.
Q does not matter for this induced charge.
 
  • #5
Is this method correct that I am using? Or should I use an image method here?
 
  • #6
Your method is not incorrect, but it involves an infinite series. I would try the image method first.
 

FAQ: Force on a point charge due to a sphere

1. What is the formula for calculating the force on a point charge due to a sphere?

The formula for calculating the force on a point charge due to a sphere is F = k * q * Q / r^2, where k is the Coulomb's constant (9x10^9 N m^2/C^2), q is the charge of the point charge, Q is the charge of the sphere, and r is the distance between the point charge and the center of the sphere.

2. How does the distance between the point charge and the center of the sphere affect the force?

The force between a point charge and a sphere is inversely proportional to the square of the distance between them. This means that as the distance increases, the force decreases.

3. Can the force on a point charge due to a sphere be attractive or repulsive?

Yes, the force can be either attractive or repulsive depending on the charges of the point charge and the sphere. If they have opposite charges, the force will be attractive, and if they have the same charge, the force will be repulsive.

4. How does the charge of the point charge and the sphere affect the force?

The force between a point charge and a sphere is directly proportional to the product of their charges. This means that as the charges increase, the force increases as well.

5. Can the shape and size of the sphere affect the force on a point charge?

Yes, the shape and size of the sphere can affect the force on a point charge. However, as long as the charge is uniformly distributed on the surface of the sphere, the force will only depend on the distance and the charges, not the shape or size of the sphere.

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