Answer:
Q_enclosed = 1.576 nC
Explanation:
Given:
- The edge length of the cube L = 1.9 m
- One corner of the cube P_1 = ( 4.8 , 13.9 ) m
- The Electric Field vector is given by:
E = 3.4 i + 4.4*y^2 j + 3.0 k N/C
Find:
What is the net charge contained by the cube?
Solution:
- The flux net Ф through faces parallel to y-z plane is:
net Ф_yz = E_x . A . cos (θ)
Where, E_x is the component of E with unit vector i.
θ is the angle between normal vector dA and E.
Hence,
net Ф_yz = 3.4 . 1.9^2 . cos (0) + 3.4 . 1.9^2 . cos (180)
net Ф_yz = 3.4 . 1.9^2 - 3.4 . 1.9^2 . cos (180)
net Ф_yz = 0.
- Similarly, The flux net Ф through faces parallel to x-y plane is:
net Ф_xy = E_z . A . cos (θ)
Where, E_z is the component of E with unit vector k.
net Ф_xy = 3 . 1.9^2 . cos (0) + 3 . 1.9^2 . cos (180)
net Ф_xy = 3 . 1.9^2 - 3 . 1.9^2 . cos (180)
net Ф_xy = 0
-The flux net Ф through faces parallel to x-z plane is:
net Ф_xz = E_y . A . cos (θ)
Where, E_y is the component of E with unit vector j.
net Ф_xz = 4.4y_1^2 * 1.9^2 . cos (0) + 4.4y_2^2. 1.9^2 . cos (180)
Where, The y coordinate for face 1 y_1 = 3.9 - 1.9 = 2, & face 2 y_2 = 3.9
net Ф_xz = - 4.4*2^2*1.9^2 . cos (0) - 4.4*3.9^2. 1.9^2 . cos (180)
net Ф_xz = -63.536 + 241.59564 = 178.0596 Nm^2/C
- From gauss Law we have:
Total net Ф_x,y,z = Q_enclosed / ∈_o
Where,
Q_enclosed is the charge contained in the cube
∈_o is the permittivity of free space = 8.85*10^-12
Hence,
Total net Ф_x,y,z = net Ф_xz + net Ф_yz + net Ф_xy
Total net Ф_x,y,z = 178.0596 + 0 + 0 = 178.0596 Nm^2/C
We have,
Q_enclosed = Total net Ф_x,y,z * ∈_o
Q_enclosed = 178.0596 * 8.85*10^-12
Q_enclosed = 1.576 nC