PHYS222_Prelab_Quiz_11_Capacitor_Lab
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Prelab Quiz 11 Capacitor Lab
Name:
Learning goals:
Students will understand the concepts of electric potential and build simple circuits with capacitors. A capacitor stores energy as an electrical field. When an electric potential is applied across a capacitor, Positive and negative charges accumulate on each side of a capacitor until the capacitor's electric potential is equal to the applied voltage. We can describe this by a simple linear equation.
Q
=
C V
(Q: Charge collected on each side, V: electric potential applied, C: capacitance)
In this quiz, we will investigate the characteristics of a capacitor. First, we will investigate the Capacitor Lab simulation. (https://phet.colorado.edu/en/simulations/capacitor-
lab )
1.
Go to the "Introduction" tab when you start the simulation. 2.
Select all options on the menu and increase the battery voltage up to +1.5 V.
A capacitor connected to a battery
3.
Increase the plate area and observe the changes in the capacitance, plate charge, stored energy, and voltage across the capacitor. Then, fill out the blanks below. (0.8 pts)
a.
As the area increases, the capacitance ______increase______.
(Reasoning
: C is proportional to A/d, C proportional to A)
b.
As the area increases, the plate charge ______increase______. (Reasoning
: Q = CV, C proportional to A, Q proportional to A)
c.
As the area increases, the stored energy ______increase______.
(Reasoning
: U is proportional to E^2, E proportional to Q, Q proportional to A, U proportional to A^2)
Or U is proportional to ½ (Q^2/C), Q^2 > C, U is proportional to Q)
d.
As the area increases, the electric potential_____the same_______.
(Reasoning
: Q = C * V => V = Q/C => => V is proportional to Q and inversely proportional to C => Q increase and C increase => V stays the same)
4.
Decrease the distance between plates and observe the changes in the capacitance, plate charge, stored energy, and voltage across the capacitor. Then, fill out the blanks below. (0.8 pts)
a.
As the distance decreases, the capacitance ______increase______.
(Reasoning
: C is proportional to A/d, C inversely proportional to d)
b.
As the distance decreases, the plate charge ______increase______.
(Reasoning
: Q = CV, C inversely proportional to d, Q inversely proportional to d)
c.
As the distance decreases, the stored energy ______increase______.
(Reasoning
: U is proportional to E^2, E proportional to Q, Q inversely proportional to A, U inversely proportional to d^2
Or U is proportional to ½ (Q^2/C), Q^2 > C, U is proportional to Q)
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d.
As the distance decreases, the electric potential _____the same_______.
(Reasoning
: Q = C * V => V = Q/C => => V is proportional to Q and inversely proportional to C => Q increase and C increase => V stays the same)
5.
From steps 3 and 4, you observed that the plate charge changes when you change the plate area or the
distance between charges. How can the plate collect more charges even though the voltage does not change? (0.5 pts)
Q = C * V, Since V is the same, we have to increase the capacitance for the Charge to increase. Since C is proportional to A/d, we either increase the area of the plate, or decrease the distance between the plate.
A capacitor charged and disconnected from a battery
6.
Click reset all and bring everything back to step 2. (Set the voltage to +1.5V again.) Then click the "Disconnect Battery" button. In the following steps, we will study the behavior of a charged capacitor
without a battery.
7.
Increase the plate area and observe the changes in the capacitance, plate charge, stored energy, and voltage across the capacitor. Then, fill out the blanks below. (0.8 pts)
a.
As the area increases, the capacitance ______increase______.
(Reasoning
: C is proportional to A/d, C proportional to A)
b.
As the area increases, the plate charge ______no change______.
(Reasoning
: No battery, no current, no change in charge that is currently on the plate)
c.
As the area increases, the stored energy _____decrease_______.
(Reasoning
: U is proportional to ½ (Q^2/C), no change in Q but C increase, U is inversely proportional to C)
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d.
As the area increases, the electric potential ______decrease______.
(Reasoning
: Q = C * V => V = Q/C => V is proportional to Q and inversely proportional to C => No change in Q but C increase => V decreases)
8.
Decrease the distance between plates and observe the changes in the capacitance, plate charge, stored energy, and voltage across the capacitor. Then, fill out the blanks below. (0.8 pts)
a.
As the distance decreases, the capacitance ______increase______.
(Reasoning
: C is proportional to A/d, C inversely proportional to d)
b.
As the distance decreases, the plate charge ______no change______.
(Reasoning
: No battery, no current, no change in charge that is currently on the plate)
c.
As the distance decreases, the stored energy _____decrease_______.
(Reasoning
: U is proportional to ½ (Q^2/C), no change in Q but C increase, U is inversely proportional to C)
d.
As the distance decreases, the electric potential______decrease______.
(Reasoning
: Q = C * V => V = Q/C => V is proportional to Q and inversely proportional to C => No change in Q but C increase => V decreases)
9.
From steps 7 and 8, you observed that the plate electric potential changes when you change the plate area or the distance between charges. How can the voltage across the capacitor charge even though the plate charge does not change? (0.5 pts)
Q = C * V => V = Q/C => V is proportional to Q and inversely proportional to C => No change in Q => V changes as C changes => C increases V decreases, C decreases V increases. The change in C depends on the Area of the plate or the distance between the plate, where C is proportional to A and inversely proportional to d.
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Series capacitor circuit
10. Go to the "Multiple Capacitors" tab. 11. Select all options and select the "2 in series" option. Choose C
1
= 1
×
10
−
13
F, and C
2
= 2
×
10
−
13
F. This
setup is identical to the LAB 2 first-part setup.
12. Measure the voltage across each capacitor and fill out the table below. (0.4 pts)
Capacitor
The voltage across a capacitor(V)
C
1
1
C
2
0.5
13. What is the equivalent capacitance of the series capacitors? (0.4 pts)
We have: Q = C * V => C = Q / V
V tot or V battery = V1 + V2 = 1 + 0.5 = 1.5 = 3/2 (V)
Q = 1 * 10 ^ -13 (C)
C equivalent = Q / V = 1 * 10 ^ - 13 / (3 / 2) = (2 / 3) * 10 ^ - 13
C equivalent = 0.7 *10 ^ -13 = 0.7 * 10 ^ - 1 *10 ^ -12 = 0.07 * 10 ^ - 12 (F) (Check, same answer as in the the capacitor lab simulation)
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