Al P. answered • 05/23/20
Online Mathematics tutor
a) increases (relative permittivity of dielctric > 1, which is true for the vast majority of (all?) dielectrics).
b) charge remains the same (series capacitors all experience the same current, and that delivers the charge Q to its plates).
c) potential difference decreases. Q=VC, so V = Q/C. Thus if Q remains the same and C increases, then V must decrease.
d) potential energy is QV/2 ==> energy will be lower because V decreases and Q remains constant.
This should be enough to figure out what happens to the other capacitor... As a strong hint, only (c) and (d) will be affected on the other capacitor.
Al P.
That's a good point. I think what happens in reality is that as the first capacitor is reconfigured, the voltage across the reconfigured capacitor starts to drop, while the voltage begins to rise across the 2nd one (the sum of the two drops must equal that of the battery). Now since i = C*dV/dt (and i = dQ/dt) ultimately Q does change somewhat, thus Q rises on that 2nd capacitor. C is a property of the physical configuration of the capacitor, it is therefore a constant for the 2nd capacitor. The only thing left that can change is V, and it must increase across the 2nd capacitor to keep C = Q/V constant.05/23/20
Dorsa R.
So if Q will rise what will happen to the fact that these are in series and Q must be the same for both ?05/23/20
Rajai A.
Hi In these problems we start with the effect of increasing the capacity of one capacitor on the total capacity of the circuit . Here C total will increase then we calculate the new total charge which will be the same for the two capacitor also q will increase ,q2=c2 ×v battery ,Because the battery is still connected . The potential deference of the capacitor with more capacitiy will be fewer ( in series) because q = c1×v1 = c2×v2 . the potential energy will be more in the capacitor with less capacity U = q square /2c . Inverse relation because q square is constant for the two capacitor . To be more confident assume that you have two series circuits . The first one with 2 mic f two capacitors and 4v battery , the second with 2,3 micro f two capacitors and 4 v battery , and solve for q1q2 ,V1 V2 , U1 U2.05/23/20
Al P.
I agree with tutor Rajai, it is good to pick a concrete example. This circuit can be modeled by the battery (Vb) connected to two series capacitors (say they are identical for simplicity: C1 = C2) with a switch connecting a 3rd identical capacitor (C3=C1=C2=C) in parallel to one of the two. The initial case has the switch open so Vc = (1/2)Vb and Qtotal = Q1=Q2 = (1/2)Vb*C. Now close the switch, Ctotal is now 1/((1/(2C)) +1/C) = (2/3)C and Qtotal = Q1=Q(2+3) = (2/3)Vb*C For the modified capacitor a) increases (in our example it is now 2C) b) increases (in our example it goes from (1/2)(Vb*C) to (2/3)(Vb*C) c) decreases (goes from (1/2)Vb to (2/3)Vb) d) decreases (goes from (1/8)C*Vb^2 to (1/9)C*Vb^2) For the unmodified capacitor a) stays the same b) increases (goes from (1/2)Vb*C to (2/3)Vb*C) c) increases (goes from (1/2)Vb to (2/3)Vb) d) increases (goes from (1/8)C*Vb^2 to (2/9)C*Vb^2) This seems like a charge pump circuit example....05/24/20
Dorsa R.
thanks for helping me05/24/20
Dorsa R.
thanks for the answer but wont the C of the second one decrease ?i know that Q remains the same but then V will increase so due to Q=CV , C must decrease right ?05/23/20