Work done by electric field

[663697]

During charging, four microcoulombs of negative charge are transferred from one plate of an 8 uF capacitor to the other plate. How much work was done by the electric field during this charging process?

a. -4 uJ
b. -1 uJ
c. 1 uJ
d. 4 uJ

Why is it necessary to use E = (1/2)*q^2*C for this question? Since it is asking for the work done, wouldn't it be more appropriate to use W = q^2 * C? Can somebody please explain in which situations either equation is more appropriate?

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[Lawpy]

I'm not sure whether the formulas you're using have the right units. Here are the SI units for each of the quantities in the energy equation.

Energy = E = joule (J)
Charge = q = coulomb (C)
Capacitance = C = farad (F)

1 farad = 1 coulomb / volt, where 1 volt (V) = 1 joule / coulomb. In other words:

1 F = 1 C / V -->
1 F = 1 C * V^-1 -->
1 F = 1 C * (J / C)^-1 -->
1 F = 1 C * (C / J) -->
1 F = 1 C^2 / J -->
1 J = 1 C^2 / F

Translating units into variables, we find that E = k * (q^2 / C), where k is some constant. When the capacitor is charging, the electric field isn't constant and changes with time. The 1/2 term comes from calculating an integral of the overall charging process.

Now the energy from the battery is E = q*V, so you're wondering why only 1/2 of the energy goes to the capacitor. This happens because the other 1/2 of the energy is irreversibly lost to heat via the resistor. You could argue that lowering the resistance should recover some of the energy loss but that doesn't happen. Remember Ohm's Law and the equations for resistance:

V = I*R and P = I*V --> P = I^2 * R, where P = power, I = current and R = resistance. Even if you decrease the resistance, increasing the charging rate in the capacitor will increase the current through the resistor (and quite rapidly due to the I^2 term), because current = charge / time = charging rate.

See the following for more information: Energy Stored on a Capacitor

 

[663697]

I'm not sure whether the formulas you're using have the right units. Here are the SI units for each of the quantities in the energy equation.

Energy = E = joule (J)
Charge = q = coulomb (C)
Capacitance = C = farad (F)

1 farad = 1 coulomb / volt, where 1 volt (V) = 1 joule / coulomb. In other words:

1 F = 1 C / V -->
1 F = 1 C * V^-1 -->
1 F = 1 C * (J / C)^-1 -->
1 F = 1 C * (C / J) -->
1 F = 1 C^2 / J -->
1 J = 1 C^2 / F

Translating units into variables, we find that E = k * (q^2 / C), where k is some constant. When the capacitor is charging, the electric field isn't constant and changes with time. The 1/2 term comes from calculating an integral of the overall charging process.

Now the energy from the battery is E = q*V, so you're wondering why only 1/2 of the energy goes to the capacitor. This happens because the other 1/2 of the energy is irreversibly lost to heat via the resistor. You could argue that lowering the resistance should recover some of the energy loss but that doesn't happen. Remember Ohm's Law and the equations for resistance:

V = I*R and P = I*V --> P = I^2 * R, where P = power, I = current and R = resistance. Even if you decrease the resistance, increasing the charging rate in the capacitor will increase the current through the resistor (and quite rapidly due to the I^2 term), because current = charge / time = charging rate.

See the following for more information: Energy Stored on a Capacitor

Yep, you're right, I meant to write E = 0.5*(q^2)/C.

The reason I'm confused about the use of the 0.5 factor is because in another similar question, they only used W = q*V rather than 0.5*q*V:

A potential difference of 10 V is present between the plates of a capacitor. How much work must be done to move 6.25 * 10^18 electrons from the positive plate to the negative plate.

Answer: 10 J

I know there has to be some sort of discrepancy between the 2 situations that would warrant using the 0.5 factor in one and not the other, but I'm having a hard time picking it out.

 

[Lawpy]

Yep, you're right, I meant to write E = 0.5*(q^2)/C.

The reason I'm confused about the use of the 0.5 factor is because in another similar question, they only used W = q*V rather than 0.5*q*V:

A potential difference of 10 V is present between the plates of a capacitor. How much work must be done to move 6.25 * 10^18 electrons from the positive plate to the negative plate.

Answer: 10 J

I know there has to be some sort of discrepancy between the 2 situations that would warrant using the 0.5 factor in one and not the other, but I'm having a hard time picking it out.

The potential difference for this question is fixed: it's a constant of 10 V. So the equation of E = q*V can be used. But in cases where the potential difference or the electric field is changing (like charging a capacitor), the equation becomes E = (1/2)*q*V.

 

[663697]

The potential difference for this question is fixed: it's a constant of 10 V. So the equation of E = q*V can be used. But in cases where the potential difference or the electric field is changing (like charging a capacitor), the equation becomes E = (1/2)*q*V.

Ohh that's what I was looking for Thanks!