Potential Drop across capacitor

Find the potential drop across capacitor $C_1$

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#Physics #ElectricityAndMagnetism #PhysicsProblem

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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Comments

$\frac{C_{2}(E_{1} +E_{2})}{C_{1} + C_{2}}$

jatin yadav - 7 years, 10 months ago

Swap the positions of E2 and C2. You now have a simple voltage divider.

Jimmy Kariznov - 7 years, 10 months ago

swapping them won't make a difference?

can you please explain a bit more?

hemang sarkar - 7 years, 10 months ago

nice writing :D

Rafael Muzzi - 7 years, 10 months ago

In steady state, charges stored in the capacitors must be same. Then, we just have to apply Kirchoff's Law.

Sambit Senapati - 7 years, 10 months ago

Not only at steady state , but at all time instants charges stored in capacitors would be same.

jatin yadav - 7 years, 10 months ago

yes, of course.

Sambit Senapati - 7 years, 10 months ago

how? can you explain a bit more?

Advitiya Brijesh - 7 years, 10 months ago

@Advitiya Brijesh – $q_{cap.} = \int\limits_0^ti_{cap.}(t) dt$ . Since , $i_{cap.}(t)$ is same for both capacitors ( as the circuit is complete ) , $q_{cap.}$ would also be same for both at any instant .I have assumed that initial charges of both capacitors is zero. (else the question would have been tidious and answer would have been a function of time.)

jatin yadav - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$