Problem 3. Consider the two large parallel plates separated by air shown below.
ID: 1682750 • Letter: P
Question
Problem 3. Consider the two large parallel plates separated by air shown below. Let each plate have a cross- sectional area of 0.01 m2 and let the voltage difference between the plates be 100 V. Assume that the separation between the plates is 0.001 m.?V
+Q | |-Q
|q |
A) Find the capacitance of the parallel plates.
B) What charge Q is stored on the plates.
C) What energy is stored in the capacitor?
D) What is the electric field in the region between the plates?
E) If a proton (charge q = 1.6 × 10-19 C, mass = 1.67 · 10-27 kg) is released from rest near the positive plate, how fast will it be moving just before it reaches the negative plate?
Explanation / Answer
Given that ,Area of a parallal plates is A = 0.01 m^2
seperation between two plates d = 0.001m
Potential difference V = 100 V
a) Capacitor of a parallel plate capacitor C = _0 A / d
= 8.85*10^-12 * 0.01 / 0.001
= 88.5 *10^-12 F
= 89 pF
b ) Charge in a Capacitor Q = CV = 89 *10 ^-12F * 100 V = 8.9*10 ^-9 C c ) Energy stored in a Capacitor U = 1 / 2 CV^2 = 1 /2 * 89* 10^-12F * ( 100V)^2 = 0.44 *10^-6 J d ) Electric field between the plates is E = V / d = 100 V / 0.001m = 1*10^5 V /m e ) charge of a proton is q = 1.6*10^-19C K.E = q V = 1.6*10^-19C *100V = 1.6*10^-17J K.E = 1/ 2 mv^2 1.6*10^-17J = 1/ 2 * 1.67 *10^-27 kg * v^2 speed of the proton v = 1.38*10^5 m/s
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