I have a question about the conservation of charge

[nksunmoon]

I have a question about the conservation of charge. the picture of the circuit is in the file uploaded.i know that the total charge of the c1 and c2 is changed after the v1 changed to v2. So the following eqution is right:Vd2-Vd1=C1（V2-V1）/（C1+C2）,we can call this equation "e1" for convenience.
Now i will list out another eqution:（Vd1-V3）*C2 + （Vd1-V1）*C1=（Vd2-V3）*C2 + （Vd2-V2）*C1,we can call this equation "e2".And we can derive "e1" from "e2",so this proves that the "e2" is also right indirectly.
But i don't know why "e2" is right. Is there any charge conservation principle in the "e2"? I have been confused by this problems for long time, is there anybody who can help me solve this problem ? Thank you very much!

 

[jimito13]

Hello nksunmoon,

The proof of the two equations is quite simple if you consider the charge conservation principle along with the definition of the capacitance for a classical parallel plate electrical capacitor,that is dQ=CdV.So let's work with your symbols along with those of the picture you uploaded :

From the charge conservation principle :

Qfinal_state=Qinitial_state => QC1_final+QC2_final=QC1_initial+QC2_initial =>

C1*(Vd1-V1)+C2*(Vd1-V3)=C1*(Vd2-V2)+C2*(Vd2-V3) => (eq. e2 proved)

=>C1*Vd1-C1*V1+C2*Vd1-C2*V3=C1*Vd2-C1*V2+C2*Vd2-C2*V3 =>

=>C1*Vd1-C1*V1+C2*Vd1=C1*Vd2-C1*V2+C2*Vd2=>

=>(C1+C2)*Vd1-C1*V1=(C1+C2)*Vd2-C1*V2=>

=>(C1+C2)*(Vd2-Vd1)=C1*(V2-V1)=>

=>(Vd2-Vd1)=(C1/C1+C2)*(V2-V1) (eq. e1 proved)

Regards,
Jimito13

 

[nksunmoon]

Hello nksunmoon,

The proof of the two equations is quite simple if you consider the charge conservation principle along with the definition of the capacitance for a classical parallel plate electrical capacitor,that is dQ=CdV.So let's work with your symbols along with those of the picture you uploaded :

From the charge conservation principle :

Qfinal_state=Qinitial_state => QC1_final+QC2_final=QC1_initial+QC2_initial =>

C1*(Vd1-V1)+C2*(Vd1-V3)=C1*(Vd2-V2)+C2*(Vd2-V3) => (eq. e2 proved)

=>C1*Vd1-C1*V1+C2*Vd1-C2*V3=C1*Vd2-C1*V2+C2*Vd2-C2*V3 =>

=>C1*Vd1-C1*V1+C2*Vd1=C1*Vd2-C1*V2+C2*Vd2=>

=>(C1+C2)*Vd1-C1*V1=(C1+C2)*Vd2-C1*V2=>

=>(C1+C2)*(Vd2-Vd1)=C1*(V2-V1)=>

=>(Vd2-Vd1)=(C1/C1+C2)*(V2-V1) (eq. e1 proved)

Regards,
Jimito13

Thank you very much for your explaination,but i don't think the total charge is conservative.You see ,the c1 and c2 are connected in series, so the total capacitance will be c1*c2/(c1+c2).When the v1 is changed to v2,the total charge of the serial capacitance is changed by (v2-v1)*c1*c2/(c1+c2).So the total charge is not conservative.According to the statement above, i think that "QC1_final+QC2_final" is not equal to "QC1_initial+QC2_initial ", then why the equation "C1*(Vd1-V1)+C2*(Vd1-V3)=C1*(Vd2-V2)+C2*(Vd2-V3) " is still right ??

 

[luben111]

Hi,

If you change V1 to V2 there will be some charge transfer between both capacitors and new voltage will appear on the middle point. So we can't speak about conservation of the charge but about redistribution of the charge.

 

[chuackl]

Let assume V3>V1 and V3>V2 also. The both equation also correct however the equation 2 does not show conservation of charge. When V3>V1 and V3>V2 is obvious V3>Vd1>V1 and V3>Vd2>V2 also. Vd1-V3 will result in a negative value. However if we consider the total charge store in the capacitor we do not minus it out(correct me if i am wrong). For equation 2 you can only say conservation of charge if Vd1-V3 have a modulus sign. The total charge in figure 1 is (V3-Vd1)C2+(Vd1-V1)C2 and figure 2 is (V3-Vd2)C1+(Vd2-V1)C2 which is not same as equation 2. You can only say it is conserve when（V3-Vd1）*C2 + （Vd1-V1）*C1=（V3-Vd2）*C2 + （Vd2-V2）*C1 but not when（Vd1-V3）*C2 + （Vd1-V1）*C1=（Vd2-V3）*C2 + （Vd2-V2）*C1

 

[dedalus]

Hi,

i don't think the total charge is conservative

When calculating total charge we must encounter charge of both plates of a capacitor.

Charge conservation principle for an isolated system: In isolated system total charge is always constant.

then why the equation "C1*(Vd1-V1)+C2*(Vd1-V3)=C1*(Vd2-V2)+C2*(Vd2-V3) " is still right ??

Node between two caps is isolated, so net (total) charge at this node must be constant . Q1 - charge of the upper plate of C1, Q2 - charge of the upper plate of C2, then:

(-Q1) + Q2 = const,

or

Q2(t2) - Q1(t2) = Q2(t1) - Q1(t1).

If we substitute Q1(t1) = (V1-Vd1)*C1, Q1(t2) = (V2-Vd2)*C1, Q2(t1) = (Vd1-V3)*C2, Q2(t2) = (Vd2-V3)*C2 in the last equation, we'll get:

(Vd1-V3)*C2 - (V1-Vd1)*C1 = (Vd2-V3)*C2 - (V2-Vd2)*C1.

 

[chuackl]

hmm sorry for making the above mistake, i confuse it with parallel capacitor calculation. In series capacitor, the charge store in C1=C2 correct? I guess what
dedalus mention is correct. Thank you dedalus.By the way, I can only figure that for figure 1 charge store in capacitor C1 minus charge store in capacitor C2 will result in 0? Same goes to figure 2. So (V3-Vd1)C2-(Vd1-V1)C1=0 and (V3-Vd2)C2-(Vd2-V3)C1 equal to 0 also. Is it possible we say that (V3-V1)C2-(Vd1-V1)C1=(V3-Vd2)C2-(Vd2-V3)C1 also since both will result in a 0 from this equation can get back the equation above also?Consider second case where V2>V3, (V3-V1)C2-(Vd1-V1)C1=(V2-Vd2)C1-(Vd2-V3)C2 Is it true to consider such a way or i just get the equation coincidently from mathematically expression ?

 

[dedalus]

I can only figure that for figure 1 charge store in capacitor C1 minus charge store in capacitor will result in 0?

Not in general. Initial amounts of charge of C1 and C2 can be different.

 

[chuackl]

When doing the calculation, isn't we consider the capacitor have been totally charged to full capacity?

 

[luben111]

Hi,

The formulas you discolsed above are correct but they can lead to completely wrong results and conclusions. In fact in the middle point between C1 and C2 you can have any voltage you want and the voltage on Figure 1 should ne considered like initial state. You need to remember that for capacitors the initial state (initial voltage on their plates) has important role. The voltage calculated on Figure 1 is correct only in case that both capacitors were fully discharged and then a voltage V1 was applied when a charge trasnfer occurs between C1 and C2.

Notice that initially Vd1 could be for example -12V and then applying V1 will result in completely different voltage. People often mistake capacitors with resistors where the final voltage doesn't depend on the initial state.

 

[chuackl]

Oh ok. Thank you very much that is something new i learned today. From a reference book i read, i only saw how the author derive the formula for capacitor. It say V=V1+V2 where V1 and V2 is the voltage across the capacitor so V=Q/C
than Q/C=Q1/V1+Q2/V2 than since it is series Q1=Q2=Q also hence Q can be cancel become equivalent capacitance1/C=1/C1+1/C2. It does not explain in detail again thank you people very much for sharing something new to me

 

[nksunmoon]

oh i see , i think that the total charge of the bottom plate of the c1 and the upper plate of the c2 is conservative,and is zero.so the equation "e2" is right ,and equals to zero.