IMAGE NOT AVAILABLE, PLEASE SHOW STEPS IF POSSIBLE: A q = 30 nC particle of mass
ID: 2239341 • Letter: I
Question
IMAGE NOT AVAILABLE, PLEASE SHOW STEPS IF POSSIBLE: A q = 30 nC particle of mass m = 167 mg is suspended at rest between the plates of a parallel plate capacitor. If the distance between the plates is 5 cm and the angle ? between the vertical and the light string the particle is suspended from is 12.0 degrees, The electric field is in he positive x-direction. a) What is the potential difference between the plates? b) What is the area charge density ? on each plate? c) If the area of each plate is 321 cm2, what is the capacitance of this capacitor?Explanation / Answer
sum forces in y
T cos theta = m g
T = mg/cos theta
sum forces in x
T sin theta = q E
m g tan theta = q E
but E = V/d
E = m g d tan(theta)/q = 167E-6*9.81*5.0E-2*tan(12)/30.0E-9=580.375
b) E = V/d
C = e0A/d
Q = CV
Q = e0 A/d V
V = sigma d/e0
E = sigma/e0
sigma = e0 E = 8.85E-12*580.375=5.14E-9
c) C = e0 A/d = 8.85E-12*321.0E-4/5.0E-2=5.68E-12 F