Capacitance and electric potential

In summary, we have a 100 pico Farad capacitor C_1 that is initially charged to a potential difference of 50 volts. After the charging battery is disconnected, C_1 is connected to a second uncharged capacitor C_2. When the potential across C_1 drops to 35 volts, we can use the equations C = Q/V and Q_1 + Q_2 = 5*10^-9 coulombs to find the capacitance of C_2. Through substitution and solving, we find that C_2 = 2.43*10^-10 Farads.
  • #1
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Homework Statement



a 100 pico Farad capacitor C_1 is charged to a potential difference of 50 volts. the charging battery is disconnected and the capacitor is then connected to a second uncharged capacitor C_2. if the potential across C_1 drops to 35 volts, what was the capacitance of C_2?


Homework Equations



capacitance C = Q/V where Q is charge, and V is electric potential in volts
units of capacitance is Farad

The Attempt at a Solution


let C_1 = Q/V_1 where C_1 = 100 pico farads = 100 *10^-12 farads and V_1 = 50 volts
solve for Q to get Q = 5*10^-9 coulombs

let C_2 = Q/V_2 where Q = 5*10^-9 coulombs and V_2 = 35 volts, solve for C_2 to get

capacitance of C_2 = 1.43*10^-10 Farads

did i do this correctly?

thanks
 
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  • #2
Be careful! Does all the charge initially on the first capacitor move onto the second?
 
  • #3
no you are right, all the charge of C_1 does not go to C_2. did soling the C_1 equation for q come into play for this type of problem?

how do i set up an equation that involves C_1 and C_2, better yet the change in capacitance of C_1?

or would the equation become C_2 = Q/ deltaV where Q = 5*10^-9 and V = 50 -35 = 25 so that C_2 = 2*10^-10 farads?

is this the correct approach?
 
  • #4
No, your right that C_2 has 35 V across it. Since the two capacitors are in parallel they must have the same voltage across them.

You know that after the battery is removed: Q_1+Q_2=5nC

Is there any way you can find Q_1 and then find Q_2?

HINT: The capacitances of capacitors are "built in quantities." Thus, C_1 is the same before and after the battery is removed.
 
  • #5
didn't i already find Q_1 when i did this calculation:
" let C_1 = Q/V_1 where C_1 = 100 pico farads = 100 *10^-12 farads and V_1 = 50 volts
solve for Q to get Q = 5*10^-9 coulombs " ??

also, how do you know that the capacitors are in parallel? if it is parallel, capacitance C = C_1 + C_2 and C_1 = 100 pico farads, how is the change in voltage factored in?
 
  • #6
scholio said:
didn't i already find Q_1 when i did this calculation:
" let C_1 = Q/V_1 where C_1 = 100 pico farads = 100 *10^-12 farads and V_1 = 50 volts
solve for Q to get Q = 5*10^-9 coulombs " ??

also, how do you know that the capacitors are in parallel? if it is parallel, capacitance C = C_1 + C_2 and C_1 = 100 pico farads, how is the change in voltage factored in?

Well, this is a situation where, in some sense, the capacitors can be said to be in parallel and they can be said to be in series so it's a bit confusing.

Bu the key point is that they are the only two elements in a loop, right? By Kirchoff's voltage law, the potential difference across each capacitor must be equal (in absolute value) so there is a 35 volts potential difference on each of them

The Q_1 you found (using the 50 volts) gives the total charge on th etwo capacitors after they have been connected. After they are connected Q on capacitor 1 decreases and Q on capacitor two increases so you now have Q_1 + Q_2 = 5*10^-9 coulombs

So you know the sum of the two charges. And you know that for each capacitor you have

Q_1 = C_1 V and Q_2 = C_2 V

(using the fact that they have the same potential difference V).
So you have three equations fo rthree unknowns (Q_1,Q_2 and C_2). So you can solve
 
  • #7
i am confused about the value of V, is it 35 or 50 volts?
once of found V is this how i would go about finding capacitance C_2:

since Q_1 + Q_2 = 5*10^-9 and Q_1 = C_1V and Q_2 = C_2V :

if i let Q_1 = 5*10^-9 - Q_2
----> sub that in for Q_1 so that 5*10^-9 - Q_2 = C_1V
----> solve that eq for Q_2 so that Q_2 = -C_1V + 5*10^-9
----> sub that into the 'Q_2 = ...' eq and then solve that so that finally

C_2 = (-C_1V + 5*10^-9)/V

did i calculate for C_2 correctly?
 
Last edited:
  • #8
In the part where the battery is connected, it is V is 50 volts. In the second part, if one capacitor has 35 volts on it, the other must have 35 volts on it, as nrqed said.

You equation for C_2 looks good to me.
 
  • #9
so for solving for C_2 using the equation above, i should use V=35volts, C_1 = 100 picocoulombs? and get:

C_2 = (-C_1V + 5*10^-9)/V = [(100*10^-12)(35) + 5*10^-9]/35 = 2.43*10^-10 farads?

correct now?
 
  • #10
scholio said:
so for solving for C_2 using the equation above, i should use V=35volts, C_1 = 100 picocoulombs? and get:

C_2 = (-C_1V + 5*10^-9)/V = [(100*10^-12)(35) + 5*10^-9]/35 = 2.43*10^-10 farads?

correct now?

A minor point: I think you dropped the minus sign in front of C_1V.
 
  • #11
okay, other than the sign, my calculation is correct?

thanks for your help
 

What is capacitance and how is it related to electric potential?

Capacitance is a measure of an object's ability to store electric charge. It is directly related to the electric potential, or voltage, between two conductors. The higher the capacitance, the more charge an object can store at a given potential.

What factors affect the capacitance of an object?

The capacitance of an object is affected by the area of the conductors, the distance between them, and the type of material between them. A larger area, closer distance, and higher permittivity material will result in a higher capacitance.

How is capacitance calculated?

The capacitance of an object can be calculated using the formula C = Q/V, where C is capacitance, Q is charge, and V is voltage. It can also be calculated using the formula C = εA/d, where ε is the permittivity of the material, A is the area of the conductors, and d is the distance between them.

What is the unit of capacitance?

The unit of capacitance is the farad (F), which is equal to one coulomb per volt. It is named after Michael Faraday, a scientist who made important contributions to the study of electricity.

What are some real-world applications of capacitance?

Capacitance is used in a variety of electronic devices, such as capacitors, which are commonly used to store energy and filter out unwanted signals. It is also used in touch screens, radio receivers, and power factor correction in electrical systems.

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