Respuesta :
Answer:
a) The Electric Field acts vertically upwards in positive direction with magnitude E = 23.9456 KN / C
b) The Electric Field acts vertically upwards in positive direction with magnitude E = 27.048 KN / C
Explanation:
Given:
- Length of non conducting wire L = 0.075 m
- Charge Density = 130*10^-9 C / m
- Point P = ( 0 , 0.055)
Find:
a) magnitude and direction of the electric field this wire produces at a point P.
Solution:
- Using , λ = q / L we note that the (infinitesimal) charge on an element dx of the rod contains charge dq = λ dx. Electric field E at point P, which is a distance r from the element is as follows:
dE = k*dq / r^2
- By symmetry, we conclude that all horizontal field components cancel and we need only “sum” (integrate) the vertical components. By symmetry , we will just integrate over half length 0 < x < L /2, and then double the results as follows:
[tex]E = 2*\frac{1}{4*pi*e_o} \int\limits {\frac{ sin(Q)}{r^2} } \, dq[/tex]
- we note that sin Q = R/r where = sqrt (^ 2 + ^2). Hence ,
[tex]E = \frac{λ*R}{2*pi*e_o}\int\limits{\frac{dx}{(x^2 + R^2)^1.5} } \,[/tex]
- Now we integrate the above equation:
[tex]E = \frac{λ*R}{2*pi*e_o}*{\frac{x}{R^2*\sqrt{(x^2 + R^2)} } } \,[/tex]
- Evaluate the Limits 0 < x < L / 2:
[tex]E = \frac{λ*R}{2*pi*e_o}*{\frac{L/2}{R^2*\sqrt{((L/2)^2 + R^2)} } } \,\\\\E = \frac{λ}{2*pi*e_o}*{\frac{L}{R*\sqrt{(L^2 + 4R^2)} } } \,[/tex]
- Use the above expression to evaluate E at point P = ( 0 , 0.0550 ):
[tex]E = \frac{130*10^-9}{2*pi*8.85*10^-12}*{\frac{0.075}{0.055*\sqrt{(0.075^2 + 4*0.055^2)} } } \, \\\\E = 2337.869 * 10.24248\\\\E = 23945.6 N/C[/tex]
Answer: The Electric Field acts vertically upwards in positive direction with magnitude E = 23.9456 KN / C .
b) Wire circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 5.50 cm directly above its center. Electric field is direct upward electric field is direct downward?
- By symmetry, we conclude that all perpendicular field components cancel and we need only integrate the vertical components.
- While the parallel component dE* cos (Q) , where cos (Q) is as shown:
cos (Q) = z / r
cos (Q) = z / sqrt (z^2 + R^2)
- Using , λ = dq / ds we note that the (infinitesimal) charge on an element ds of the circular wire contains charge dq = λ ds. Electric field E at point P, which is a distance r from the element is as follows:
[tex]dE = \frac{1}{4*pi*e_o}*\frac{λ ds}{r^2} \\\\dE = \frac{1}{4*pi*e_o}*\frac{λ ds}{z^2 + R^2} \\\\dE*cos(Q) = \frac{λ}{4*pi*e_o}*\frac{z ds}{(z^2 + R^2)^1.5}\\\\[/tex]
- Now we integrate over the entire length of the circular wire 0 < s < 2piR
[tex]E= \frac{λL}{4*pi*e_o}*\frac{z}{(z^2 + L^2 / 4*pi^2)^1.5}\\\\\\[/tex]
- Compute Electric Field at point P:
[tex]E= \frac{130*10^-9*0.075}{4*pi*8.85*10^-12}*\frac{0.055}{(0.055^2 + 0.075^2 / 4*pi^2)^1.5}\\\\E = 87.67 * 308.5256\\\\E = 27.048KN/C[/tex]
Answer: The Electric Field acts vertically upwards in positive direction with magnitude E = 27.048 KN / C .
