Answer:
Net charge = 2.59nC
Explanation:
Gauss' Law states that the net electric flux is given by:
∮→E⋅→d/A = q enc/ϵ0
At this point, we have to solve the net electric flux through each side.
One corner is at (4.8, 3.9, 0).
The other corners are each 1.9m apart, so the other corners are at
(4.8, 5.8, 0), (6.7, 3.9, 0) and (6.7, 5.8, 0).
The other four corners are 1.9m away in the z-axis:
(4.8, 5.8, 2), (6.7, 3.9, 2) and (6.7, 5.8, 2).
Therefore, there are two planes (parallel to the x-z plane), one at
y = 3.9
and the other at
y = 5.8 which have a constant y-coordinate and are facing in the −^j and +^j respectively.
Area = 1.9m * 1.9m = 3.61m²
The flux through the first plane (area of 3.61m²) is given by:
E.A = (3.4i + 4.4 * 3.9²j + 3.0k) * (-3.61m²j) = -241.59564
The flux through the other plane is
E.A = (3.4i + 4.4 * 5.8²j + 3.0k)* (3.61m²j) = 534.33776
Now, for the other planes. There are no ^j components for the unit vectors for the area.
Therefore, even though they change in y-coordinate, those terms cancel out.
Therefore, for the planes with unit vector in the x-direction, we get:
E.A = (3.4i + 4.4y²j + 3.0k) * (1.9m * 1.9m) = ±12.274
And in the z-direction:
E.A = (3.4i + 4.4y²j + 3.0k) * (1.9m * 1.9m) = ±10.83
Now, when we sum all these fluxes together, the contribution from the x- and z-directions cancel out. Therefore, our net flux is:
-241.59564 + 534.33776 = 292.74212
Therefore, the enclosed charge is given by
q enc = ϵ0* (292.74212)
= 2.5856871136363E−9C
= 2.59E-9 nC--_- Approximated
= 2.59nC
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
