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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 734: Lesson Check

Exercise 44

Step 1

1 of 1

The charge of a capacitor is directly proportional to the voltage of the capacitor. The relation is given by

$$
Q=CV
$$

Where $C$ is the capacitance of the capacitor.

Exercise 45

Step 1

1 of 2

We start by the relationship $PE = QV/2$, which gives an expression for the energy stored in a capacitor with charge $Q$ and potential difference $V$. Suppose the capacitor has a definite capacitance $C$, then we have $Q = CV$, or $V = Q/C$. In such case, the energy stored in the capacitor can be expressed in terms of $Q$ as

$$
PE = frac{1}{2}Qleft(frac{Q}{C}right) = frac{Q^2}{2C}
$$

where $C$ is constant.

Result

2 of 2

$$
PE = dfrac{Q^2}{2C}
$$

Exercise 46

Step 1

1 of 2

An electric field exerts an electric force on an electric charge, causing it to move. In this way the electric field does work on the electric charge.

The mechanical work done to separate two unlike charges is stored in the electric field as electric potential energy. This energy can be converted into kinetic energy when the charges are set free.

Result

2 of 2

An electric field does work on an electric charge. The mechanical work is stored in the electric field as electric potential energy.{}

Exercise 47

Step 1

1 of 2

Capacitance is a characteristic property of all capacitors, independent of the amount of charge on their plates. So even when the capacitor is uncharged, the capacitance is still there, not zero.{}

Result

2 of 2

no; capacitance is a nonzero-valued, characteristic property of the capacitor, independent of its state of charging{}

Exercise 48

Step 1

1 of 2

If the plate separation is decreased, the electric field between the plate will increase, hence the attractive potential will also increase. Hence we can hold more charge with less voltage applied. Hence the capacitance will increase.

Result

2 of 2

Capacitance will increase.

Exercise 49

Step 1

1 of 2

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

$$
begin{align*}
V &= frac{Q}{C}\
&= frac{56times10^{-6};mathrm{C}}{750times10^{-6};mathrm{F}}\
&= 0.075;mathrm{V}
end{align*}
$$

Result

2 of 2

0.075 V{}

Exercise 50

Step 1

1 of 2

The required capacitance can be calculated using the relationship $Q = CV$. Solving for $C$ and substituting the known values, we find

$$
begin{align*}
C &= frac{Q}{V}\
&= frac{32times10^{-6};mathrm{C}}{9.0;mathrm{V}}\
&= 3.6;mumathrm{F}
end{align*}
$$

Result

2 of 2

$3.6;mu$F

Exercise 51

Step 1

1 of 2

The energy stored in the device is given by $PE = QV/2$. But we know that $Q = CV$, so we can write the potential energy as

$$
PE = frac{1}{2}(CV)V = frac{1}{2}CV^2
$$

Solving for $C$ and substituting the known values, we find

$$
begin{align*}
C &= frac{2PE}{V^2}\
&= frac{2(125;mathrm{J})}{(1050;mathrm{V})^2}\
&= 227;mumathrm{F}
end{align*}
$$

Result

2 of 2

$227;mu$F

Exercise 52

Step 1

1 of 2

$textbf{(a)}$ Using $Q = CV$, we find

$$
begin{align*}
Q &= CV\
&= (890times10^{-6};mathrm{F})(330;mathrm{V})\
&= 0.294;mathrm{C}
end{align*}
$$

$textbf{(b)}$ The energy stored in the flash unit is

$$
begin{align*}
PE &= frac{1}{2}QV\
&= frac{1}{2}(0.294;mathrm{C})(330;mathrm{V})\
&= 48.5;mathrm{J}
end{align*}
$$

Result

2 of 2

$textbf{(a)}$ 0.294 C $textbf{(b)}$ 48.5 J

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