Q69P A1.50鈭抦聽cylinder of radius 1.... [FREE SOLUTION]

91影视

University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 4: Q69P (page 845)

A $1 .50 鈭� m$ cylinder of radius $1.10 cm$ is made of a complicated mixture of materials. Its resistivity depends on the distance $x$ from the left end and obeys the formula $蚁 (x) = a + {bx}^{2}$ , where aand bare constants. At the left end, the resistivity is $2 .25 脳 10^{鈭� 8} 惟$ m, while at the right end it is localid="1668398251768" $8 .50 脳 10^{鈭� 8} 惟鈰� m$ . (a) What is the resistance of this rod? (b) What is the electric field at its midpoint if it carries 1.75 - Acurrent? (c) If we cut the rod into two 75.0 - cmhalves, what is the resistance of each half?

Short Answer

Expert verified

(a) The resistance of the rod is $1.71 脳 10^{鈭� 4} 惟$

(b) The electric field at its midpoint is $1.76 脳 10^{4} V / m$

Step by step solution

Determine the resistance of the rod

We know, $蚁 (0) = a$ , so

$\begin{aligned} 蚁 (0) = a = 2.25 脳 10^{鈭� 8} 惟鈰� m \\ 蚁 (L) = 8.50 脳 10^{鈭� 8} 惟鈰� m \\ A n d, 蚁 (L) = 2.25 脳 10^{鈭� 8} 惟鈰� m + b (1.50 m)^{2} \end{aligned}$

So, for b localid="1668398625029" $\begin{aligned} b = \frac{8.50 脳 10^{鈭� 8} 惟鈰� m 鈭� 2.25 脳 10^{鈭� 8} 惟鈰� m}{(1.50 m)^{2}} \\ b = 2.78 脳 10^{鈭� 8} 惟 / m \end{aligned}$

Now, cross-sectional area of the cylinder:

$\begin{aligned} A = 蟺 r^{2} \\ A = 蟺 (0.011 m)^{2} \\ A = 3.80 脳 10^{鈭� 4} m^{2} \end{aligned}$

Therefore,

$\begin{aligned} R = \frac{(2.25 脳 10^{鈭� 8} 惟鈰� m) (1.50 m)}{3.80 脳 10^{鈭� 4} m^{2}} + \frac{(2.78 脳 10^{鈭� 8} 惟 / m) (1.50 m)^{3}}{3 (3.80 脳 10^{鈭� 4} m^{2})} \\ R = 1.71 脳 10^{鈭� 4} 惟 \end{aligned}$

Thus, the resistance of the rod is $1.71 脳 10^{鈭� 4} 惟$

Determine the electric field at the midpoint of the rod

$\begin{aligned} E = 蚁 J \\ W e k n o w, E = (a + b x^{2}) \frac{I}{A} \end{aligned}$

At midpoint of the cylinder

$E = \frac{l (a + b L^{2} / 4)}{A}$

Substitute the values in the equation

$\begin{aligned} E = \frac{(1.75 A) [(2.25 脳 10^{鈭� 8} 惟鈰� m) + (2.78 脳 10^{鈭� 8} 惟 / m) (1.50 m)^{2} / 4]}{3.80 脳 10^{鈭� 4} m^{2}} \\ E = 1.76 脳 10^{4} V / m \end{aligned}$

Therefore, the electric field at the midpoint of the cylinder is

Determine the resistance of each half

For the left half:

Integrate dR in the limits of $x = 0 鈫� L / 2$

$\begin{aligned} R_{L H} = 鈭� d R = \frac{1}{A} {鈭�}_{0}^{L / 2} 鈥� (a + b x^{2}) d x \\ R_{L H} = \frac{1}{A} {[a x + \frac{b x^{3}}{3}]}_{0}^{L / 2} \\ R_{L H} = \frac{a L}{2 A} + \frac{b L^{3}}{24 A} \end{aligned}$

Substitute the given values

$\begin{aligned} R_{L H} = \frac{(2.25 脳 10^{鈭� 8} 惟鈰� m) (1.50 m)}{2 脳 (3.80 脳 10^{鈭� 4} m^{2})} + \frac{(2.78 脳 10^{鈭� 8} 惟 / m) (1.50 m^{3})}{24 (3.80 脳 10^{鈭� 4} m^{2})} \\ R_{L H} = 5.47 脳 10^{鈭� 5} 惟 \end{aligned}$

For the right half:

To determine the resistance of the right half, subtract the resistance of left half from the total resistance. So,

$\begin{aligned} R_{R H} = R 鈭� R_{L H} \\ R_{R H} = 1.71 脳 10^{鈭� 4} 惟鈭� 5.47 脳 10^{鈭� 5} 惟 \\ R_{R H} = 1.16 脳 10^{鈭� 4} 惟 \end{aligned}$

Therefore, the resistance of the left half of the cylinder is $5.47 脳 10^{鈭� 5} 惟$ , and the resistance of the right half of the cylinder is $1.16 脳 10^{鈭� 4} 惟$