Two 10.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. T

Question

Two 10.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. The electrodes are connected to the terminals of a 15 V battery. What are (i)the electric field inside the capacitor, (ii)the charge on each electrode, and (iii)the potential difference between the electrodes:

(a) while the capacitor is attached to the battery and

(b)After insulating handles are used to pull the electrodes from each other until they are 1.0 cm apart? The electrodes remain connected to the battery during this process.

Explanation / Answer

Hi,

In this case we have a capacitor formed by two plates with the shape of a circle of known radius. To answer this question we must do something first.

If we think about the distance between the plates (d) and the size of them, the first is much smaller than the second (in both cases (a) and (b) ) so we can assume that the field between the parallel plates is uniform.

Besides, we should consider two things: that the potential difference between the plates and the potential difference of the terminals of the battery are equal, and that the charge of both plates has the same magnitude but opposite sign.

So, the only thing we should remember is how to calculate the electric field and the charge in this situation.

V = Ed ; where V is the potential difference, E is the electrical field and d is the distance between the plates.

E = Q / (A*e0) ; where e0 is the vacuum permittivity which value is 8.85*10-12 C2/Nm2 and Q is the charge on either plate.

(a) In this part we have the following:

(i) E = V/d = (15 V)/(0.5*10-2 m) = 3000 N/C

(ii) Q = E*A*e0 = (3000 N/C)**( 5*10-2 m)2*(8.85*10-12 C2/Nm2) = 2.08*10-10 C

(iii) V = 15 V

(b) For this part we have the following:

(i) E = V/d = (15 V)/(1*10-2 m) = 1500 N/C

(ii) Q = E*A*e0 = (1500 N/C)**( 5*10-2 m)2*(8.85*10-12 C2/Nm2) = 1.04*10-10 C

(iii) V = 15 V

I hope it helps.

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