Does anyone know why the formula they are using is PE=qv whereas everywhere else I've looked it's given as PE=1/2 qv. This is question 11 on Princeton Review Practice Test 4--capacitors (which my physics teacher completely skipped so I am lost). Thanks!
Princeton Review Test 4 #11 on Physical Sciences Section
- Thread starter jddoc2015
- Start date
- Joined
- Apr 29, 2011
- Messages
- 2,171
- Reaction score
- 863
I don't have access to the tests but I believe the formula is
PE = QV
I could be totally wrong but I think the reason is it PE = 1/2 QV for capacitors is because the amount of work needed to charge a capacitor varies with the state of charging so 1/2QV is like an "average".
PE = QV
I could be totally wrong but I think the reason is it PE = 1/2 QV for capacitors is because the amount of work needed to charge a capacitor varies with the state of charging so 1/2QV is like an "average".
- Joined
- Jun 3, 2014
- Messages
- 272
- Reaction score
- 43
I have lost my access to the TPR tests, do you mind posting the specific question? Energy stored in a capacitor should be PE = (1/2)QV. Did they use QV in their solution?
It's passage based but here goes: Question and explanation (and passage below):
A Geiger counter is an instrument used to detect the products of radioactive decay. It consists of a cylindrical metal chamber containing a low-pressure inert gas and a wire which is connected through a resistor to the cylinder, as shown in the figure below.
Figure 1
Geiger counter
A Geiger counter operates as follows. The cylinder–gas wire assembly acts as a capacitor, setting up a strong electric field between the cylinder and the wire. When a sufficiently energetic particle of ionizing radiation enters the chamber, it ionizes the gas, thus freeing electrons. These electrons are accelerated by the electric field to such an extent that they ionize other atoms of gas. This chain reaction results in a current pulse that is transformed into a voltage pulse by the resistor. If the voltage pulse is large enough the counter will emit a click.
If the electric field within the chamber is 1 × 1010 V/m, and a newly released electron, originally at rest, travels one nanometer before colliding with a gas atom and ionizing it, what is the highest possible value for ionization energy of that atom?
A.
1 × 10–19 eV
B.
0.1 eV
C.
10 eV
Correct Answer
D.
1 × 1019 eV
Explanation:
C. The energy of the released electron would be ΔPE = qV = qEd = (1 e)(1010 V/m)(10–9 m) = 10 eV. If this electron is to ionize a gas atom, then the ionization energy of the gas can be no greater than 10 eV.
A Geiger counter is an instrument used to detect the products of radioactive decay. It consists of a cylindrical metal chamber containing a low-pressure inert gas and a wire which is connected through a resistor to the cylinder, as shown in the figure below.
Figure 1
A Geiger counter operates as follows. The cylinder–gas wire assembly acts as a capacitor, setting up a strong electric field between the cylinder and the wire. When a sufficiently energetic particle of ionizing radiation enters the chamber, it ionizes the gas, thus freeing electrons. These electrons are accelerated by the electric field to such an extent that they ionize other atoms of gas. This chain reaction results in a current pulse that is transformed into a voltage pulse by the resistor. If the voltage pulse is large enough the counter will emit a click.
If the electric field within the chamber is 1 × 1010 V/m, and a newly released electron, originally at rest, travels one nanometer before colliding with a gas atom and ionizing it, what is the highest possible value for ionization energy of that atom?
1 × 10–19 eV
0.1 eV
10 eV
Correct Answer
1 × 1019 eV
Explanation:
C. The energy of the released electron would be ΔPE = qV = qEd = (1 e)(1010 V/m)(10–9 m) = 10 eV. If this electron is to ionize a gas atom, then the ionization energy of the gas can be no greater than 10 eV.
- Joined
- Apr 29, 2011
- Messages
- 2,171
- Reaction score
- 863
@jddoc2015
The question is asking about how much energy needed to move an electron. That's why you use PE = QV.
The PE = 1/2QV is the energy stored in a capacitor.
I think the "V" when calculating PE of the capacitor comes from the battery (whose voltage is not given). The "V" when calculating the PE of the electron comes from the "V=Ed" and they give you E and d.
The question is asking about how much energy needed to move an electron. That's why you use PE = QV.
The PE = 1/2QV is the energy stored in a capacitor.
I think the "V" when calculating PE of the capacitor comes from the battery (whose voltage is not given). The "V" when calculating the PE of the electron comes from the "V=Ed" and they give you E and d.
