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10Q

Expert-verifiedFound in: Page 794

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

*Cap-monster maze***. **

The charge on capacitor C is $60\mathrm{\mu C}$.

$\mathrm{V}=10\mathrm{V}\phantom{\rule{0ex}{0ex}}\mathrm{C}=6.0\mathrm{\mu F}$

**Here, use the formula for the charge in terms of the voltage and the capacitor. When the capacitors are connected in series with battery, voltage remains same through the capacitors. **

Formulae are as follow:

$\mathrm{q}=\mathrm{CV}$

Where, *C* is capacitance,* V* is potential difference, *q* is charge

Consider the loop as shown in figure 27-22 in which the capacitor C is connected in series with the battery shown by the path in red.

So, the voltage across capacitor C is 10 V

Now, the charge is as follow,

$\mathrm{q}=\mathrm{CV}\phantom{\rule{0ex}{0ex}}\mathrm{q}=6.0\times 10\phantom{\rule{0ex}{0ex}}\mathrm{q}=60\mathrm{\mu C}$

Hence, the charge on capacitor C is $60\mathrm{\mu C}$

** **

Therefore, first find the voltage through the capacitors. From that, find the charge through the capacitors.

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