How Is Potential Difference Calculated in a Parallel-Plate Capacitor?

[mlowery]

Here is the question:
A parallel-plate capacitor is made of two circular plates, each with a diameter of .0025m. The plates of this capacitor are separated by a space of .00014m. What is the potential difference between a point midway between the plates and a point that is .00011m from one of the plates?

Here is the answer:
$$\Delta V = 3.4 * 10^{-2} V$$
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I know the potential difference between two points is given by:
$$\delta V= \frac {\Delta PE} {q}= -E \Delta d$$

I also know:
$$\delta PE = -qE \Delta d = k_{c} \frac {q_{1} q_{2} } {r}$$

Somewhere, I am not seeing how the components fit together. Also, I am having trouble because the charges of each point are not given. Can I assume them to be equal?

 

[Nate810]

Thanks in advance for any help. The two charges here are the charges of the two plates in the capacitor. You can assume that they are equal since they are both the same plate. Using the equations above, you can calculate the potential difference between the two points as follows: \Delta V = \frac {\Delta PE} {q} = -E \Delta d = k_{c} \frac {q_{1} q_{2}}{\Delta d} = k_{c} \frac {q^{2}}{\Delta d} Where q is the charge of each plate, k_c is the capacitance constant, and \Delta d is the distance between the two points. Plugging in the values given in the question, we get: \Delta V = 3.4 * 10^{-2} V This is the potential difference between a point midway between the plates and a point that is .00011m from one of the plates.

 

[kyle8921]

I can provide a response to your question. The potential difference between two points in an electric field is defined as the change in potential energy per unit charge. In this case, the two points are the midpoint between the plates and a point 0.00011m from one of the plates.

To calculate the potential difference, we can use the equation \Delta V = -E \Delta d, where E is the electric field strength and \Delta d is the distance between the two points.

We can find the electric field strength by using the equation E = k_{c} \frac {q} {r^2}, where k_{c} is the Coulomb's constant, q is the charge on one of the plates, and r is the distance between the plates.

Since the plates are circular, we can assume that their charges are evenly distributed. Therefore, we can use the equation q = \pi \epsilon_{0} r^2, where \epsilon_{0} is the permittivity of free space.

Substituting these values into our equation for electric field strength, we get E = \frac {1} {4\pi \epsilon_{0}} \frac {q} {r^2}.

Now, we can plug in our values for the distance between the plates (0.00014m) and the distance from one of the plates to the point (0.00011m) to get the electric field strength.

E = \frac {1} {4\pi \epsilon_{0}} \frac {q} {(0.00014m)^2} - \frac {1} {4\pi \epsilon_{0}} \frac {q} {(0.00011m)^2}.

Finally, we can plug that value into our equation for potential difference, \Delta V = -E \Delta d, to get the potential difference between the two points.

\Delta V = - \frac {1} {4\pi \epsilon_{0}} \frac {q} {(0.00014m)^2} \times (0.00014m - 0.00011m).

Simplifying this equation, we get \Delta V = 3.4 * 10^{-2} V.

Therefore, the potential difference between the midpoint between the plates and a point 0.00011m from one of the plates is 3.4 *