Consider the figure below that shows the capacitor connected to a 10 volt DC sou
ID: 2304503 • Letter: C
Question
Consider the figure below that shows the capacitor connected to a 10 volt DC source. The capacitor has a value of 2.0 microfarads.
Determine the charge stored on the capacitor in microCoulombs.
Charge stored on capacitor: _________microCoulombs
Now, using one of the equations given in the document, determine the energy stored within the capacitor, in Joules.
Energy stored within the capacitor: ________Joules
Now, referring to the next figure (below), with the battery still connected, please slide a piece of dielectric material with a dielectric constant of 2.0 between these plates.
Determine the new capacitance, the new charge, the new voltage across the capacitor, and the new energy stored within the capacitor.
New capacitance: Cnew = __________microFarads
New voltage across the capacitor Vnew = _________ volts
New charge on the capacitor Qnew = ________ microCoulombs
New energy stored by the capacitor Enew = _______ Joules
Finally, as shown in this last figure, imagine that you disconnect the capacitor from the battery and THEN insert the dielectric material.
Determine the new capacitance, the new charge, the new voltage across the capacitor, and the new energy stored within the capacitor for this case.
New capacitance: Cnew = ________ microFarads
New charge on the capacitor Qnew = _______ microCoulombs
New voltage across the capacitor Vnew = ________ volts
New energy stored by the capacitor Enew = ____________ Joules
Explanation / Answer
a)
Charge stored on the capacitor
Q=CV =2*10 =20 uC
b)
Energy stored in the capacitor
E=(1/2)CV2 =(1/2)*(2*10-6)*102 =1*10-4 J
c)
New Capacitance
Cnew=KC =2*2u =4 uF
Voltage across the capacitor
Vnew=10 V
New charge on the capacitor
Qnew=4u*10 =40 uC
Energy stored on the capacitor
Enew =(1/2)*(4*10-6)*102 =2*10-4 J
d)
New Capacitance
Cnew=KC =2*2u =4 uF
New charge on the capacitor
Qnew =20 uC (Since charge remains constant)
New voltage across the capacitor
Vnew =Qnew/Cnew =20/4 =5 V
New energy stored by the capacitor
Enew =(1/2)*(Q2/C) =(1/2)(20*10-6)2/(4*10-6)
Enew=5*10-5 J
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