Question

A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacitor?

Answer 1

voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

$c = \frac{c1 \times c2}{c1 + c2}$

$= \frac{2 \times 4}{2 + 4}$

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

$= 1.33 \times 10 {}^{ - 6} \times 8$

$= 10.64 \times 10 {}^{ - 6}$

charge on 2.0μf capacitor is

$\frac{Qeq}{2 \times 10 {}^{ - 6} }$

$= \frac{10.64 \times 10 {}^{ - 6} }{2 \times 10 {}^{ - 6} }$

=5.32v

learn more about series capacitance from here: https://brainly.com/question/28166078

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