Question

Hi,

Can you please help me solve this problem? It tells me i'm off by a factor of 10 so i tried putting 258 and 25800 for part a and -258 and -25800 for part b but it kept saying incorrect... Can you please find the mistake and solve this question all over again for me?

Thanks alot!

1. 1/3 points | Previous Answers SerPSE7 24.P.004. My Notes Ask Your Teacher Consider a closed triangular box resting within a horizontal electric field of magnitude E = 8.60 x 104 N/C, as shown in the figure below. 30 cm 0 c60 10 cm (a) Calculate the electric flux through the vertical rectangular surface of the box -2580 Your response is off by a multiple of ten. kN-m2/c (b) Calculate the electric flux through the slanted surface of the box 2580 Your response is off by a multiple of ten. kN-m2/c (c) Calculate the electric flux through the entire surface of the box Need Help? Talk to a Tutor

Answer 1

answer ) a) we have the formula

$\Phi$=$\oint$E.dA=$\oint$E.dAcos$\Theta$=$\oint$E.dAcos180^o=-E$\oint$dA= -EA= -8.60*10⁴N/C*(0.30m*0.10m=-2.580 kN-m²/C

so the answer to part a is -2.58 kN-m²/C or -2.580 kN-m²/C ( here we have taken the value of of the angle 180 because the electric field and the area is opposite to one another )

b) again we have

$\Phi$=$\oint$E.dA=$\oint$E.dAcos$\Theta$=EAcos60^o

from the diagram below we have

cos60^o=0.10m/L

L=0.10/cos60^o

putting the values in above equation we have

$\Phi$=8.60*10⁴*0.10/cos60*0.30*cos60=2.580kN-m²/C

so answer to part b is 2.58kN-m²/C or 2.580kN-m²/C

c) the flux through entire surface of the box will be zero because the dA vector is acting perpendicular away from the interior of the box

so cos$\Theta$=90^o=0

so $\Phi$( entire surface)=0