Why is the Charge on the Equivalent Capacitor the Same as C1?

[Strawberry]

Homework Statement

Q17. Capacitors C1 and C2 are connected in series and a potential difference is applied to the combination. If the capacitor that is equivalent to the combination has the same potential difference, then the charge on the equivalent capacitor is the same as:

A. the charge on C1

Homework Equations

q = CV
1/Ceq = sum ( 1/Ci )

The Attempt at a Solution

Well the answer is given, but I don't understand it. I keep doing the problem according to the formulas, but I get q1*q2 / q1+q2

 

[alphysicist]

No, that does not look right. I think the place where you are going wrong is maybe you are thinking that $q_1=C_1 V$. (This is not true because C1 does not have the full battery voltage V across it--part of it is across C1 and part across C2.) What is true is that $q_1 = C_1 V_1$.

So what is the charge on the equivalent capacitance in terms of C1,C2, and V?

Once you have that, you can find the charge on C1 by itself.

 

[Moetasim]

= q1.

As a scientist, it is important to approach problems methodically and logically. In this case, we can use the given equations to understand the problem and come up with a solution.

First, let's define the variables:

C1 and C2 are the two individual capacitors in series
Ceq is the equivalent capacitance of the series combination
q1 and q2 are the charges on C1 and C2, respectively

Now, we can use the equation q=CV to relate charge, capacitance, and potential difference. In this case, since the potential difference is the same for both capacitors, we can write:

q1 = C1V and q2 = C2V

Next, we can use the equation 1/Ceq = sum (1/Ci) to find the equivalent capacitance. In this case, since there are only two capacitors in series, we can write:

1/Ceq = 1/C1 + 1/C2

Finally, we can substitute our expressions for q1 and q2 into the equation q1*q2 / q1+q2 = q1 to find the charge on the equivalent capacitor. This gives us:

qeq = q1*q2 / q1+q2 = (C1V)*(C2V) / (C1V + C2V) = (C1C2V^2) / (C1+C2)

So, the charge on the equivalent capacitor is given by (C1C2V^2) / (C1+C2). This is not the same as q1, as given in the answer. This may be a mistake in the problem or the answer key. It is important to always check our calculations and results to ensure they make sense and are consistent with the given information.

In conclusion, as a scientist, it is important to approach problems systematically and critically evaluate the results we obtain. In this case, we can use the given equations and definitions to understand the problem and come up with a solution, but we should also double-check our calculations and results to ensure they are accurate and make sense in the context of the problem.