A capacitor consists of two charged disks of radius 3.2 m separated by a distanc
ID: 2055541 • Letter: A
Question
A capacitor consists of two charged disks of radius 3.2 m separated by a distance s = 2 mm (see the figure). The magnitude of the charge on each disk is 43 µC. Consider points A, B, C, and D inside the capacitor, as shown in the diagram. The distance s1 = 1.3 mm, and the distance s2 = 0.5 mm. (Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.)
(a) What is the electric field inside the capacitor?
E =? N/C
(b) First, calculate the potential difference VB - VA.
What is Delta l along this path?
Delta l =? m
What is VB - VA?
VB - VA = ? volts
(c) Next, calculate the potential difference VC - VB.
What is Delta l along this path?
Delta l = ? m
(d) Finally, calculate the potential difference VA - VC.
What is Delta l along this path?
Delta l = ? m
What is VA - VC?
VA - VC = ? volts
please use vector notation where applicable thanks =)
Explanation / Answer
Energy stored in a capacitor is 0.5 * C * V^2. Charge in a capacitor is C * V. a) Capacitance is eo * er * A / d We can see that the difference in capacitance will be 6.7/1.00054 times the air capacitor. The voltage difference will be 1.2E7 / 3E6. Using the energy formula, we see that energy is proportional to capacitance and V^2. So multiply the air cap's energy by the voltage difference squared times the capacitance difference. You should end up with E = 0.086J * (1.2E7/3E6)^2 * (6.7/1.0054)
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