Two 3.0-cm-diameter aluminum electrodes are spaced 0.50 mm apart. The electrodes are connected to a 100 V battery.

a. What is the capacitance?

b. What is the magnitude of the charge on each electrode?

Short Answer

Expert verified

(a) The value of the capacitance isC=12.5pF.

(b) The magnitude of the charge on each electrode isQ=2.5nC.

Step by step solution

01

Part (a) Step 1: Given Information

We need to define the capacitance.

02

Part (a) Step 2: Explanation

The capacitance of this type of parallel plate capacitor is given by

C=ε0Ad

where isAthe area of the plate anddis the distance between the plates. In this case

A=D24π

Where,D=3.0cmis the diameter of the plate. This further given as

C=ε0(D2π)4d

Using the numerical values and we get

C=12.5pF

03

Part (b) Step 1: Given Information

We need to define the magnitude of the charge on each electrode.

04

Part (b) Step 2: Explanation

On each electrode, the magnitude of the charge is given by

Q=CΔV=12.5pF100V=2.5nC

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