CAPA set 3
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1. The potential difference between two parallel conducting plates in vacuum is
205 V. An alpha particle with mass of 6.50×10-27 kg and charge of 3.20×10-19
C is released from rest near the positive plate. What is the kinetic energy of the
alpha particle when it reaches the other plate? The distance between the plates
is 45.0 cm.: QV=energy (J)
2. Three capacitors of capacitance C1=4.50 μF, C2 =5.00 μF, and C3=10.0 μF are
connected to a 34.0 V battery as shown in the figure.
(a) Calculate the charge on C3.
(b) Calculate the voltage across C1.: (a) Ceq=C3(C1+C2)/C3+(C1+C2)
Q=CeqV
(b) V across C1=Q/C
3. A parallel-plate air capacitor of area A= 23.0 cm2 and plate separation d= 2.40
mm is charged by a battery to a voltage 54.0 V. If a dielectric material with κ =
3.50 is inserted so that it fills the volume between the plates (with the capacitor
still connected to the battery), how much additional charge will flow from the
battery onto the positive plate?: C=Aε0/d= (23e-4)(8.85e-12)/d
Q=CV
C`=κ(Aε0/d)
Q`=C`V
Q0=Q`-Q
4. Two capacitors C1 = 3.8 μF, C2 = 14.5 μF are charged individually to V1 = 13.2
V, V2 = 3.2 V. The two capacitors are then connected together in parallel with
the positive plates together and the negative plates together.
(a) Calculate the final potential difference across the plates of the capacitors
once they are connected.
(b) Calculate the amount of charge (absolute value) that flows from one capac-
itor to the other when the capacitors are connected together.
(c) By how much (absolute value) is the total stored energy reduced when the
two capacitors are connected?: (a) Q1i=C1V1, Q2i=C2V2
Qf=Q1i+Q2i
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Vf=Qf/(C1+C2)
(b)Q1=C1V1, Q2=C2V2
Q1+Q2=Qeq
q1/C1=q2/C2 --> Qeq-q2/C1=q2/C2
find q1, q2
q2-Q2=amount of charge
(c)Q1=C1V1, Q2=C2V2
Qt/(C1+C2)=Veq
C1(Veq)=q1
C2(Veq)=q2
Q1-Veq=f
[0.5(C1)(V1)^2]+[0.5(C2)(V2)^2]=e1 (J)
0.5(C1+C2)(Veq)=e2 (J)
e1-e2=total energy reduced (J)
5. In the section of circuit below, R1 1.650 Ω, R2 1.780 Ω, and R3 0.810 Ω. Find
the equivalent resistance of this combination of resistors.: R2, R3 series
R1, R2R3 parallel
R2+R3=R23
1/R=1/R23+1/R1=1/equivalent resistance
6. In the section of circuit below, R1= 3.426 Ω, R2= 1.060 Ω, and the voltage
difference Va-Vb= 1.000 V. The current i= 0.250 A.
(a) Find the value of R3.
(b) What is the current through R3.: (b) V across R1=R1I
V across R2+R3=1-(V across R1)
I through R2=(V across R2+R3)/R2
I through R3=I-(I through R2)
(b) R3=(V across R2+R3)/(I through R3)
2/8
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1. The potential difference between two parallel conducting plates in vacuum is
205 V. An alpha particle with mass of 6.50×10-27 kg and charge of 3.20×10-19
C is released from rest near the positive plate. What is the kinetic energy of the
alpha particle when it reaches the other plate? The distance between the plates
is 45.0 cm.: QV=energy (J)
2. Three capacitors of capacitance C1=4.50 μF, C2 =5.00 μF, and C3=10.0 μF are
connected to a 34.0 V battery as shown in the figure.
(a) Calculate the charge on C3.
(b) Calculate the voltage across C1.: (a) Ceq=C3(C1+C2)/C3+(C1+C2)
Q=CeqV
(b) V across C1=Q/C
3. A parallel-plate air capacitor of area A= 23.0 cm2 and plate separation d= 2.40
mm is charged by a battery to a voltage 54.0 V. If a dielectric material with κ =
3.50 is inserted so that it fills the volume between the plates (with the capacitor
still connected to the battery), how much additional charge will flow from the
battery onto the positive plate?: C=Aε0/d= (23e-4)(8.85e-12)/d
Q=CV
C`=κ(Aε0/d)
Q`=C`V
Q0=Q`-Q
4. Two capacitors C1 = 3.8 μF, C2 = 14.5 μF are charged individually to V1 = 13.2
V, V2 = 3.2 V. The two capacitors are then connected together in parallel with
the positive plates together and the negative plates together.
(a) Calculate the final potential difference across the plates of the capacitors
once they are connected.
(b) Calculate the amount of charge (absolute value) that flows from one capac-
itor to the other when the capacitors are connected together.
(c) By how much (absolute value) is the total stored energy reduced when the
two capacitors are connected?: (a) Q1i=C1V1, Q2i=C2V2
Qf=Q1i+Q2i
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Vf=Qf/(C1+C2)
(b)Q1=C1V1, Q2=C2V2
Q1+Q2=Qeq
q1/C1=q2/C2 --> Qeq-q2/C1=q2/C2
find q1, q2
q2-Q2=amount of charge
(c)Q1=C1V1, Q2=C2V2
Qt/(C1+C2)=Veq
C1(Veq)=q1
C2(Veq)=q2
Q1-Veq=f
[0.5(C1)(V1)^2]+[0.5(C2)(V2)^2]=e1 (J)
0.5(C1+C2)(Veq)=e2 (J)
e1-e2=total energy reduced (J)
5. In the section of circuit below, R1 1.650 Ω, R2 1.780 Ω, and R3 0.810 Ω. Find
the equivalent resistance of this combination of resistors.: R2, R3 series
R1, R2R3 parallel
R2+R3=R23
1/R=1/R23+1/R1=1/equivalent resistance
6. In the section of circuit below, R1= 3.426 Ω, R2= 1.060 Ω, and the voltage
difference Va-Vb= 1.000 V. The current i= 0.250 A.
(a) Find the value of R3.
(b) What is the current through R3.: (b) V across R1=R1I
V across R2+R3=1-(V across R1)
I through R2=(V across R2+R3)/R2
I through R3=I-(I through R2)
(b) R3=(V across R2+R3)/(I through R3)
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