Question

Another Capacitance Question

Answer 1

Two identical capacitors Ca and Cb of capacitance 4.20 µF are connected in series across a total potential of 570 V. A dielectric slab of dielectric constant 3.30 can fill Ca and is slowly inserted into that capacitor.

What is the change in charge, ΔQa, on Ca when the dielectric is added to Ca?

I understand the theory for this question but when it comes to devising the equation properly.
Clearly a C=Q/V question but I don't know how to set it up right. I know the K(dielectric constant)=E/E' and all that comes into play its simply organizing my thoughts

Answer 2

First determine the charge on Ca without the dielectric. Since Ca = Cb then the voltage across Ca is 285 V and of course Qa=Qb.
Now the dielectric increases Ca by 3.3. The charge on each capacitor will still be the same but changed. Determined the voltage across Ca. Calculate the charge on Ca.

Answer 3

Another approach is to determine the new value of Ca with the dielectric in place. The new value will simply be 3.3 times the old value. Next using the voltage divider equation for series connected capacitors, calculate the voltage across Ca
V(Ca)= (4.2)(570)/(13.86 + 4.2) = 132.558 volt Noe use Q = CV
Compare with the original charge on Ca (use original value of 4.2 uf and the voltage of 285 volts.