What is the Capacitance of a Capacitor When a Conductor is Inserted?

[electricblue7]

Homework Statement

An isolated capacitor with capacitance C = 1 µF has a charge Q = 22 µC on its plates.
Now a conductor is inserted into the capacitor. The thickness of the conductor is 1/3 the thickness of the capacitor and is centered in between the plates of the capacitor.

What is the capacitance of the capacitor with the conductor in place?

Homework Equations

Q = C / V

V = E d

The Attempt at a Solution

I know that since the capacitor is a conductor it makes the distance between the two plates smaller which means that the capacitance should therefore increase. I thought that since we now had two distances that were each 1/3 the original distance the capacitance would increase by a factor of 6 however this answer does not work. Can anyone help me please?!?

 

[rl.bhat]

After introduction of conductor, each capacitor is three times the original capacitor. Now these two capacitors are in series. Then what will be the combined capacitor?

 

[tomcue]

I would first like to clarify that the term "conductor" in this context may refer to a material with high electrical conductivity, or it could also refer to a physical object that conducts electricity. For the purpose of this response, I will assume that it refers to a physical object.

Now, let's address the problem at hand. When a conductor is inserted into a capacitor, it changes the electric field between the plates. As you correctly stated, this results in a decrease in the distance between the plates, which in turn increases the capacitance. However, the increase in capacitance is not directly proportional to the decrease in distance between the plates.

To determine the new capacitance, we need to take into account the geometry of the conductor and its effect on the electric field. In this case, the thickness of the conductor is 1/3 of the distance between the plates. This means that the electric field will be affected by the presence of the conductor, but not completely blocked.

To calculate the new capacitance, we can use the equation C = ε0A / d, where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. In this case, the area of the plates remains the same, but the distance between them is now equal to 2/3 of the original distance. Plugging in these values, we get C = (ε0A) / (2/3)d.

Since the thickness of the conductor is 1/3 of the distance between the plates, we can also write this as C = (ε0A) / (2/3)(1/3)d. Therefore, the new capacitance is equal to 3/2 times the original capacitance.

In summary, the capacitance of the capacitor with the conductor in place is 3/2 times the original capacitance, or 1.5 µF. This is different from your initial calculation of an increase by a factor of 6, which is likely due to not taking into account the geometry of the conductor. I hope this explanation helps!