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Exercise 39

Step 1

1 of 2

The capacitance can be calculated using $Q = CV$. Solving for $C$ gives

$$
C = frac{Q}{V}
$$

Using $Q = 9.06times10^{-9}$ C (from Guided Example 20.12) and the new value of $V;(= 35.0$ V), we find

$$
C = frac{9.06times10^{-9};mathrm{C}}{35.0;mathrm{V}} = 259;mathrm{pF}
$$

Result

2 of 2

259 pF{}

Exercise 40

Step 1

1 of 2

Using $Q = CV$, we find

$$
begin{align*}
Q &= CV\
&= (150times10^{-12};mathrm{F})(12;mathrm{V})\
&= 1.8times10^{-9};mathrm{C}
end{align*}
$$

Result

2 of 2

$1.8times10^{-9}$ C

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