Q73P Electric charge is distributed u... [FREE SOLUTION]

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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 4: Q73P (page 782)

Electric charge is distributed uniformly along a thin rod of length a, with total charge Q. Take the potential to be zero at infinity. Find the potential at the following points (Fig. P23.73): (a) point P, a distance x to the right of the rod, and (b) point R, a distance y above the right颅hand end of the rod. (c) In parts (a) and (b), what does your result reduce to as x or y becomes much larger than a?

Short Answer

Expert verified

(a) Potential at the point P is $\frac{Q}{4 蟺系_{0}} \ln (\frac{x + a}{x})$

(b) Potential at the point R is $\frac{Q}{4 蟺系_{0}} \ln 鈦� (\frac{\sqrt{y^{2} + r^{2}} + a}{y})$

\frac{Q}{4 蟺 系_{0} x}

, the potential at point R when y becomes greater

\frac{Q}{4 蟺 系_{0} y}

Step by step solution

Step 1:

(a) As given,

Rod length is (a), the net charge of the rod is Q, and potential is zero at infinity.

For the point P, in order to evaluate the differential potential due to the small component, an infinitesimally small portion is taken, thus the relation is

$d V_{P} = \frac{1}{4 蟺系_{0}} \frac{d Q}{x + r}$

Now the differential charge,

$\begin{aligned} \frac{d Q}{d r} = \frac{Q}{a} \\ d Q = \frac{Q d r}{a} \end{aligned}$

Substituting the above equation,

$d V_{P} = \frac{1}{4 蟺系_{0}} \frac{\frac{Q d r}{a}}{x + r}$

Calculation

Integrating both sides:

$\begin{aligned} V_{P} = {鈭�}_{0}^{a} 鈥� \frac{1}{4 蟺系_{0}} \frac{\frac{Q d r}{a}}{x + r} \\ = \frac{Q}{4 蟺系_{0}} {鈭�}_{0}^{a} 鈥� \frac{1}{x + r} d r \\ = \frac{Q}{4 蟺系_{0} a} \ln 鈦� (\frac{x + a}{x}) \end{aligned}$

Therefore, the potential at point P is

\frac{Q}{4 蟺 系_{0}} \ln (\frac{x + a}{x})

Step 3:

(b) For the point R, in order to evaluate the differential potential due to the small component, an infinitesimally small portion is taken, thus the relation is

$d V_{R} = \frac{1}{4 蟺系_{0}} \frac{d Q}{\sqrt{y^{2} + r^{2}}}$

Now the differential charge:

$\begin{aligned} \frac{d Q}{d r} = \frac{Q}{a} \\ d Q = \frac{Q d r}{a} \end{aligned}$

Substituting the above equation;

$d V_{R} = \frac{1}{4 蟺系_{0}} \frac{\frac{Q d r}{a}}{\sqrt{y^{2} + r^{2}}}$

Integrating both side;

$\begin{aligned} V_{R} = {鈭�}_{0}^{a} 鈥� \frac{1}{4 蟺系_{0}} \frac{\frac{Q d r}{a}}{\sqrt{y^{2} + r^{2}}} \\ = \frac{\frac{Q}{a}}{4 蟺系_{0}} {鈭�}_{0}^{a} 鈥� \frac{1}{\sqrt{y^{2} + r^{2}}} d r \\ = \frac{Q}{4 蟺系_{0} a} \ln (\frac{\sqrt{y^{2} + r^{2}} + a}{y}) \end{aligned}$

Hence, the potential at the point R is

\frac{Q}{4 蟺 系_{0}} \ln (\frac{\sqrt{y^{2} + r^{2} + a}}{y})

Step 4:

(c)The potential at point P is when x becomes greater

$\begin{aligned} V_{P} = \frac{Q}{4 蟺系_{0} a} \ln 鈦� (\frac{x + a 鈭� x}{a}) \\ = \frac{Q}{4 蟺系_{0}} (\ln 鈦� (\frac{x + a}{a}) 鈭� \ln 鈦� (\frac{x}{a})) \\ = \frac{Q}{4 蟺系_{0}} \frac{1}{x} \\ = \frac{Q}{4 蟺系_{0} x} \end{aligned}$

Hence, the potential at point P when x becomes greater $\frac{Q}{4 蟺系_{0} x}$ .

Now, the potential at the point R, if y is much greater than the rod鈥檚 length;

$\begin{aligned} V_{P} = \frac{Q}{4 蟺系_{0} a} \ln (\frac{\sqrt{y^{2} + r^{2} + a}}{y}) \\ = \frac{Q}{4 蟺系_{0} a} \ln (\frac{y + a}{y}) \\ = \frac{Q}{4 蟺系_{0}} (\ln 鈦� (\frac{y + a}{a}) 鈭� \ln 鈦� (\frac{y}{a})) \\ = \frac{Q}{4 蟺系_{0}} \frac{1}{y} \\ = \frac{Q}{4 蟺系_{0} y} \end{aligned}$