Electric Potential of Spheres

[Saketh]

I have two problems that confuse me for similar reasons. The first one:

Find the potential $\varphi$ of an uncharged conducting sphere outside of which a point charge $q$ is located at a distance $l$ from the sphere's center.
​

The second one:

A system consists of two concentric conducting spheres, with the inside sphere of radius $a$ carrying a positive charge $q_1$. What charge $q_2$ has to be deposited on the outside sphere of radius $b$ in order to reduce the potential of the inside sphere to zero?​

There's more to the second problem, but this first part confused me enough.

For the first one, I originally went about it by defining the inner sphere's radius as $R$, and then using the law of cosines to find the distance between the point charge and the surface of the sphere as a function of [tex]\theta[/itex]. This, however, ended in failure, with undefined results.

Then I thought, "maybe the potential of the sphere is located at the center of the sphere?" So I wrote down $$\varphi_0 = \varphi_q + \varphi_s$$. Since $$\varphi_q$$ is $$\frac{q}{4\pi \epsilon_0 l}$$ if the potential is at the center of the sphere, and $$\varphi_s$$ is $$\frac{0}{4\pi \epsilon_0 R}$$, the potential of the sphere must be $$\frac{q}{4\pi \epsilon_0 l}$$. This is the right answer, but I am still confused - I thought the potential of a sphere should be treated as if it were on the sphere's surface, not as if it were at the center?

For the second one, I still wasn't sure if potential should be treated as surface or center, so I calculated blindly.

If the potentials are at the center, as gave me the correct answer for the first problem, then in order for the potential of the center sphere to become zero the potential of the outside sphere must cancel it out.

$$\varphi_a = -\varphi_b$$

After integrating those expressions from their differential parts, I concluded that $$\frac{-q_1 a}{b} = q_2$$, which is the wrong answer. I then tried it with Gauss's Law, but I still got the wrong answer.

I'm probably messing up because I don't understand electric potential as it applies to spheres.

In summary, I have two main questions:

1. Did I do the first problem correctly, and, if so, why is it correct?
2. How am I supposed to set up the second problem?

Thank you for your time.

 

[Saketh]

I've figured out the first question, so I don't need help with that one, but the second one (with the two spheres) still confuses me.

 

[kyle8921]

I can understand your confusion with these two problems. Electric potential of spheres can be a tricky concept to grasp, but with some explanation, I hope to clear up your confusion.

First, let's discuss the first problem. You are correct in thinking that the potential of a sphere should be treated as if it were on the surface, not at the center. This is because electric potential is a scalar quantity and is defined at a point in space, not at a particular point inside or outside of an object. So, when we talk about the potential of a sphere, we are referring to the potential at any point on its surface.

Now, to solve the first problem, you can use the equation for electric potential due to a point charge, which is \varphi = \frac{q}{4\pi\epsilon_0 r}, where r is the distance from the point charge to the point where we want to calculate the potential. In this case, r would be the distance between the point charge and the surface of the sphere. So, the potential at any point on the surface of the sphere would be \varphi = \frac{q}{4\pi\epsilon_0 l}. This is the same answer you got, but with a different approach.

Moving on to the second problem, let's break it down step by step. First, we know that the potential inside the inner sphere is not affected by the charge on the outer sphere. This is because the inner sphere is a conductor, so any charge placed on its surface will distribute itself evenly on the surface, leaving the potential inside unchanged. So, to reduce the potential of the inner sphere to zero, we need to deposit a charge on the outer sphere that will cancel out the potential due to the charge on the inner sphere.

Now, let's consider Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. In this problem, we can use a Gaussian surface that encloses only the outer sphere. Since we want the potential inside the inner sphere to be zero, we can set the flux equal to zero and solve for the charge on the outer sphere. This will give us the equation:

\frac{q_1}{\epsilon_0} = \frac{q_2}{b}

Solving for q_2, we get q_2 = \frac{q