What is the unit vector n on the bottom face of the box? n = What is the electri
ID: 1448915 • Letter: W
Question
What is the unit vector n on the bottom face of the box?
n =
What is the electric flux on the bottom face of the box?
bottom = V·m
What is the electric flux on the top face of the box?
top = V·m
What is the electric flux on the left face of the box?
left = V·m
What is the electric flux on the right face of the box?
right = V·m
What is the electric flux on the front face of the box?
front = V·m
What is the electric flux on the back face of the box?
back = V·m
How much charge is inside the box (amount and sign)?
Qinside = C
Explanation / Answer
Here,
E1 = < 60, 48, 0 > N/C
E2 = < 135, 108, 0 > N/C
for the bottom of the face of the box
unit vector n of the bottom face = 0.09 * 0.02 * m^2 (-j)
unit vector n of the bottom face = -1.8 * 10^-3 j m^2
unit vector n of the bottom face = < 0 , -1.8 * 10^-3 , 0 > m^2
-------------------
electruc flux on the bottom face = E1 * Area
electruc flux on the bottom face = < 60, 48, 0 > . < 0 , -1.8 * 10^-3 , 0 >
electruc flux on the bottom face = - 1.8 *10^-3 * 48 N.m^2/C
electruc flux on the bottom face = 0.0865 N.m^2/C
-----------------------------------
for the eletric flux through the top face = E2 * Area
the eletric flux through the top face = < 135, 108, 0 > . < 0 , 1.8 * 10^-3 , 0 >
the eletric flux through the top face = 108 * 1.8 *10^-3 N.m^2/C
the eletric flux through the top face = 0.195 N.m^2/C
the the eletric flux through the top face is 0.195 N.m^2/C
----------------------------------------------
for the electric flux through the left face = E2 * Area
the electric flux through the left face = < 135, 108, 0 > . < 0.02 * 0.09 , 0 , 0 >
the electric flux through the left face = 135 * 0.02 * 0.09 N.m^2/C
the electric flux through the left face = 0.243 N.m^2/C
the electric flux through the left face is 0.243 N.m^2/C
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.