Respuesta :
Answer:
Part I:
(a): [tex]7.889\times 10^2\ Nm.[/tex]
(b): [tex]0\ Nm.[/tex]
(c): [tex]6.52\times 10^2\ Nm.[/tex]
Part II:
(a): [tex]2.664\times 10^{-9}\ C.[/tex]
(b): Charge on the sphere is positive and only distributed on the surface of the sphere and not inside the sphere.
Explanation:
Part I:
Assuming,
[tex]\hat i,\ \hat j,\ \hat k[/tex] are the unit vectors along positive x, y and z axes respectively.
Given, the electric field is of intensity 3.22 kN/C and is along x-axis.
Therefore,
[tex]\vec E = 3.22\ kN/C\ \hat i=3.22\times 10^3\ N/C\ \hat i.[/tex]
The magnetic flux through a surface is defined as
[tex]\phi = \vec E\cdot \vec A[/tex]
where,
[tex]\vec A[/tex] is the area vector of the surface which is directed along the normal to the plane of the surface and its magnitude is equal to the area of the surface.
[tex]A = \text{length of the plane }\times \text{width of the plane}\\=0.700\times 0.350\\=0.245\ m^2.[/tex]
(a):
When the plane is parallel to yz plane, then its normal is along x axis, therefore,
[tex]\vec A = A\ \hat i.[/tex]
Electric flux,
[tex]\phi = \vec E\cdot \vec A \\=(3.22\times 10^3)\hat i\cdot (0.245)\hat i = 3.22\times10^3\times 0.245\times (\hat i\cdot \hat i) = 7.889\times 10^2\ Nm.[/tex]
(b):
When the plane is parallel to xy plane, then its normal is along z axis, therefore,
[tex]\vec A = A\ \hat k.[/tex]
Electric flux,
[tex]\phi = \vec E\cdot \vec A \\=(3.22\times 10^3)\hat i\cdot (0.245)\hat k = 3.22\times10^3\times 0.245\times (\hat i\cdot \hat k) = 0\ Nm.[/tex]
(c):
When the normal to the plane makes an angle [tex]34.2^\circ[/tex] with the x-axis.
Electric flux through the plane is given by,
[tex]\phi = \vec E\cdot \vec A\\=EA\cos(34.2^\circ) \\=(3.22\times 10^3)\times (0.245)\times \cos(34.2^\circ)\\=6.52\times 10^2\ Nm.[/tex]
Part II:
(a):
Given:
- Electric field everywhere on the surface of the sphere, [tex]E = 570\ N/C.[/tex]
- Radius of the sphere, [tex]R = 0.205\ m.[/tex]
The electric field is directed radially outward from the surface of the sphere at its every point and the normal to the surface of the sphere is also radially outwards at its every point therefore both the electric field and the area vector of the sphere is along the same direction.
The flux through the surface is given by
[tex]\phi = \vec E \cdot \vec A \\= EA\cos0^\circ\\=EA\\=E\ 4\pi R^2\\=570\times 4\pi \times 0.205^2\\=3.01\times 10^2\ Nm.[/tex]
According to Gauss law of electrostatics,
[tex]\phi = \dfrac q{\epsilon_o}[/tex]
where,
- [tex]q[/tex] = charge enclosed by the sphere.
- [tex]\epsilon_o[/tex] = electric permittivity of free space =[tex]8.85\times10^{-12}\ C^2N^{-1}m^{-2}.[/tex]
Therefore, the net charge on the sphere is given by
[tex]q=\epsilon_o\times \phi\\=8.85\times 10^{-12}\times 3.01\times 10^2\\=2.664\times 10^{-9}\ C.[/tex]
(b):
Since, the sphere is charged all of the charge resides on its surface. The electric field through the sphere is radially outwards which means the charge on the sphere should be positive.
There is no charge distributed inside the sphere.
