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?

 

[member 553083]

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.