All Solutions

Exercise 37

Step 1

1 of 2

Known:

$q=2.7times10^{-5}$ C

$V=9.0$ V

Solution:

We knwo that

$$
q=CV
$$

$$
C=frac{q}{V}=frac{left(2.7times10^{-5} {rm C}right)}{left(9.0 {rm V}right)}=3.0times10^{-6} {rm F}=3.0 {rm {rm mu F}}
$$

Result

2 of 2

3.0 textmu F

Exercise 38

Step 1

1 of 2

The potential difference of the plates can be found using $Q = CV$. Solving for $V$ and substituting the known values, we find

$$
begin{align*}
V &= frac{Q}{C}\
&= frac{5.8times10^{-6};mathrm{C}}{3.2times10^{-6};mathrm{F}}\
&= 1.8;mathrm{V}
end{align*}
$$

Result

2 of 2

1.8 V{}

unlock