A potential difference of V=80.0V is applied across two capacitors in series,with capacitances 17.0 nF and 8.00 nF.What is the charge on the 17.0 nF capacitor?

Question

Quizplus · Accepted Answer

First, we can use the capacitance formula for capacitors in series, which is:

$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$

where $C_{eq}$ is the equivalent capacitance. Plugging in the values given:

$\frac{1}{C_{eq}} = \frac{1}{17.0	imes 10^{-9}	ext{F}} + \frac{1}{8.00	imes 10^{-9}	ext{F}}$

Solving for $C_{eq}$:

$C_{eq} = 5.47	imes 10^{-9}	ext{F}$

Next, we can use the equation for the charge on a capacitor:

$Q = C	imes V$

where Q is the charge, C is the capacitance, and V is the potential difference. Plugging in the values for the 17.0 nF capacitor:

$Q = (17.0	imes 10^{-9}	ext{F})(80.0	ext{V}) = 1.36	imes 10^{-6}	ext{C}$

Therefore, the answer is C) 435 nC

A Potential Difference of V=80