A potential difference of 300 V is applied to a series connection of two capacitors of capacita

A potential difference of 300 V is applied to a series...

A potential difference of $300 \mathrm{~V}$ is applied to a series connection of two capacitors of capacitances $C_{1}=2.00 \mu \mathrm{F}$ and $C_{2}=8.00 \mu \mathrm{F}$. What are (a) charge $q_{1}$ and (b) potential difference $V_{1}$ on capacitor 1 and $(\mathrm{c}) q_{2}$ and $(\mathrm{d}) V_{2}$ on capacitor $2 ?$ The charged capacitors are then disconnected from each other and from the battery. Then the capacitors are reconnected with plates of the same signs wired together (the battery is not used). What now are (e) $q_{1}$, (f) $V_{1}$, (g) $q_{2}$, and (h) $V_{2}$ ? Suppose, instead, the capacitors charged in part (a) are reconnected with plates of opposite signs wired together. What now are (i) $q_{1},(\mathrm{j}) V_{1},(\mathrm{k}) q_{2},$ and $(\mathrm{l}) V_{2} ?$

A potential difference of $300 \mathrm{~V}$ is applied to a series connection of two capacitors of capacitances $C_{1}=2.00 \mu \mathrm{F}$ and $C_{2}=8.00 \mu \mathrm{F}$. What are (a) charge $q_{1}$ and (b) potential difference $V_{1}$ on capacitor 1 and $(\mathrm{c}) q_{2}$ and $(\mathrm{d}) V_{2}$ on capacitor $2 ?$ The charged capacitors are then disconnected from each other and from the battery. Then the capacitors are reconnected with plates of the same signs wired together (the battery is not used). What now are (e) $q_{1}$,
(f) $V_{1}$,
(g) $q_{2}$, and (h) $V_{2}$ ? Suppose, instead, the capacitors charged in part
(a) are reconnected with plates of opposite signs wired together. What now are
(i) $q_{1},(\mathrm{j}) V_{1},(\mathrm{k}) q_{2},$ and $(\mathrm{l}) V_{2} ?$

Key Concepts

Charge Redistribution in Isolated Capacitor Networks

After disconnecting a capacitor network from an external power source, the total charge within the isolated system remains conserved. However, reconnecting the capacitors in a new configuration will lead to a redistribution of this charge among the capacitors. Understanding this conservation of charge is essential to predict how voltages and charges will adjust when the circuit configuration is altered without additional charge input.

Voltage Division in Series Circuits

In a series circuit, the total applied voltage is divided among the capacitors in inverse proportion to their capacitances. Since each capacitor holds the same charge, a capacitor with a smaller capacitance will have a larger voltage drop, and vice versa. This principle is critical when calculating individual voltages across capacitors connected in series.

Effects of Reconnection Based on Plate Polarity

The way capacitors are reconnected—whether by joining plates of like sign or opposite sign—significantly affects the resulting charge distribution and potential differences across each capacitor. Connecting same-signed plates generally leads to charge sharing that equalizes potential differences following different rules compared to connecting oppositely-signed plates, where the charges may partially cancel each other. This concept is imperative for understanding dynamic behavior in reconfigured capacitor circuits.

Series Configuration of Capacitors

When capacitors are connected in series, they share the same charge even though their individual capacitances may differ. This concept is fundamental in circuit analysis because it means that despite a potentially different voltage drop across each capacitor, the amount of charge stored is identical for each capacitor in the series loop.

Capacitor Charge-Voltage Relationship

The relationship Q = CV is the cornerstone of capacitor behavior, linking the charge (Q) stored on a capacitor to its capacitance (C) and the voltage (V) across it. This proportionality allows for the determination of one variable when the other two are known, and it is used extensively in analyzing both series and parallel capacitor configurations.

Transcript

00:01 For this problem on the topic of capacitance, we are told that a potential difference of 300 volts is being applied to a series connection of two capacitors.

00:09 The capacitancees are 2 microferds and 8 microferds for c1 and c2 respectively.

00:15 We want to find the charge q1, the potential difference v1 across capacitor 1.

00:20 We want to find q2 and v2 on capacitor 2.

00:23 And we're then told that the capacitors are disconnected from each other and from the battery.

00:27 They are reconnected with the plates of same signs wired together.

00:32 We want to now find q1, v1, q2, v2, and then lastly, if the capacitors are charged in part a are reconnected with the plates of opposite signs wired together, we want to find q1, v1, q2 and v2.

00:49 Now the equivalent capacitance, c equivalent, is equal to c1, c2, divided by c1, c1.

01:03 Plus c2 for capacitors in series.

01:06 And so the charge on each capacitor, q, is equal to q1, is equal to q2, is equal to the equivalent capacitance times the potential difference v, which is c1, c2, v over c1 plus c2.

01:27 And this is two micro -farrads times 8 microfarids times the emf of 300 volts divided by 2 microferds plus 8 microferds which gives us the charge q1 and q2 to both be 4 .8 times 10 to the minus 4 couloms.

02:01 Now for part b, the potential difference v1 is equal to q over c1, which is 4 .8 times 10 to the minus 4 kuloms divided by 2 microferds, which gives the potential difference to be 240 volts.

02:31 For part c, as we noted in part a, the charge q2 is the same as q1.

02:46 And so this was calculated to be 4 .8 times 10 to the minus 4 coulums.

03:03 For part d, v2 is equal to v minus v1...

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