Question

actice Assignment Gradebook ORION Downloadable eTextbook gnment MESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VIRSION : イBACK NEXT Chapter 26, Problem 024 Your answer is partially correct. Try again. Figure (a) gives the magnitude E(x) of the electric fields that have been set up by a battery along a resistive rod of length L 9.00 mm (see Figure (b)). The vertical scale is set by E, 6.40 x10' v/m. The rod consists of three sections of the same material but with different radii. (The schematic diagram of the figure (b) below does not indicate the different radii.) The radius of section 3 is 2.15 mm. What is the radius of (a) section 1 and (b) section 2? x (mm)

Answer 1

We know that the
Electric field (E) = potential /distance = V/d
Now for the first part
$E_{1} = \frac{V_{1}}{d}$
where d is length of first part = 3 mm = 3*10^-3 m
$4*10^{3} = \frac{V_{1}}{3*10^{-3}}$
V₁ = 12 volt
Similarly for part 2
$E_{2} = \frac{V_{2}}{d}$
$6.4*10^{3} = \frac{V_{2}}{3*10^{-3}}$
V₂ = 19.2 volt
Now the third part
$2.4*10^{3} = \frac{V_{3}}{3*10^{-3}}$
V₃ = 7.2 volt
Now we know that all part are in series therefore the current must be same
I₁ = I₂ = I₃
we know that
I = V/R
where R is resistance = pL./A
where A is area = $\pi$r²
therefore
I = (V $\pi$r² /pL )
Therefore
$\frac{V_{1}R_{1}^{2}\pi }{\rho L} = \frac{V_{2}R_{2}^{2}\pi }{\rho L} =\frac{V_{3}R_{3}^{2}\pi }{\rho L}$
Since matter and length is constant therefore
$V_{1}R_{1}^{2} = V_{2}R_{2}^{2} = V_{3}R_{3}^{2}$
On putting the values , we get
$12R_{1}^{2} = 19.2*R_{2}^{2} = 7.2*2.15^{2}$
Therefore
R₁ = 1.67 mm
R₂ = 1.73 mm