What equation am I supposed to use for potential difference?

In summary, a 4.0-F capacitor has enough stored energy to heat 2.5 kg of water from 21 degree Celsius to 95 degree Celsius. The potential difference across the plates is V=-W/q. Now, you need to find out how much energy is required to heat the water right. This is given by the old thermodynamic equation: E= (m)(c)(Tf-Ti). Energy - units (joules) m - mass (units grams) c - specific heat capacity (for water, it's value is 4.186 J/(g)(K) Tf - Final temperature Ti - initial temperature
  • #1
destinee20
6
0
A huge 4.0-F capacitor has enough stored energy to heat 2.5 kg of water from 21 degree Celsius to 95 degree Celsius. What is the potential difference across the plates?

What equation am I supposed to use for potential difference?
Is it V = Change in PE / q
V= - W / q
V = (PE)(.5)/ Q

Also, I'm supposed to find how much energy is required to heat the water right? What equation do I use for that?

I'm just confused with this problem... Can somebody please help. Thanks
 
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  • #2
The amount of charge a capacitor can store is directly proportionnal to the potential difference between its plates. This constant of proportionality is the capacitance, C, meaured in Farad (F). You are told here that C = 4 F. Hence, for a capacitor plugged to a battery of potential difference [itex]\Delta V[/itex],

[tex]Q_{\mbox{final}} = C \Delta V[/tex]

Where the index "final" refers to the process during which the capacitor charges itself. At any given moment during this process, if the charge (in absolute value) on each plate is q, then the potential difference between the plates is [itex]\Delta V = q/C[/itex].

Now we contemplate that if we could find Q_final, we'd have won. Fortunately, we are given another clue: at the end of the charging process, the capacitor has enough stored energy to heat 2.5 kg of water from 21 degree Celsius to 95 degree Celsius. Figure out how much energy that it, and compare it to the energy "stored" in the capacitor. The energy stored in the capacitor is the total work done to charge the capacitor from q=0 to q=Q. (Think integral)
 
  • #3
Two things that I think can help you. First you need to find out how much energy is stored in water as it changes it's temperature. This is given by the old thermodynamic equation:

E = (m)(c)(Tf-Ti)

Energy - units (joules)
m - mass (units grams)
c - specific heat capacity (for water, it's value is 4.186 J/(g)(K)
Tf - Final temperature
Ti - initial temperature

This relation should look familiar too you. Once you get the energy from this, set it equal to one of the energy capacitance equations. The one you are probably looking for is the one that omits Q:

E = 1/2(Q)(V)

Substitute for the charge Q with CV in the above equation and you get:

E = 1/2 (C)(V)^2

Hope this helped.
 
  • #4
Thanks for the help!
 

1. What is the equation for potential difference?

The equation for potential difference is V = IR, where V represents potential difference (measured in volts), I represents current (measured in amperes), and R represents resistance (measured in ohms).

2. How do I calculate potential difference?

To calculate potential difference, you can use the equation V = IR, where V is the potential difference, I is the current, and R is the resistance. You can also use Ohm's law, which states that V = I * R, or rearrange the equation to find I or R depending on what information you have.

3. Can I use any other equations to find potential difference?

Yes, there are other equations that can be used to find potential difference, such as V = Q/C, where V is potential difference, Q is charge (measured in coulombs), and C is capacitance (measured in farads). This equation is commonly used in circuits with capacitors.

4. How is potential difference related to electric potential and electric field?

Potential difference is directly related to electric potential and electric field. Electric potential is defined as the electric potential energy per unit charge, while electric field is the force per unit charge. The potential difference between two points is equal to the electric field multiplied by the distance between those points.

5. Can I use the potential difference equation for any type of circuit?

Yes, the potential difference equation (V = IR) can be used for any type of circuit, whether it is a simple circuit with only a battery and resistor, or a more complex circuit with multiple components. However, different equations may be needed for circuits with specific components, such as capacitors as mentioned in question 3.

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