Calculation OF flux

Gauss Law Application|| Flux Due To Point Charge In Cube||Calculation OF Charge By Flux||Discussion-HW-5

Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field will be due to.

A positive charge of 17.7 muC is placed at the centre of a hollow sphere of radius 0.5m . Calculate the flux density through the surface of the sphere.

A uniform electric field of 200NC^-1 exists in space in the X-direction. Calculate the flux of this field through a plane square area of edge 10 cm placed in the Y-Z

Given the components of an electric field as E_x=alphax , E_y=0 and E_z=0 , where alpha is a dimensional constant. Calculate the flux through each face of the cube of side a, as shown in the figure.

A charge q is placed at a height (h)/(4) from the base of a cone of height h and radius R as shown in figure. Calculate the flux linked with the curved surface of the cone.

The electric field componets due to a charge inside the cube of side 0.1m are E_(x) = alpha x, where alpha = 500 (N//C) m^(-1) , E_(y) = 0, E_(z) = 0 . Calculate the flux through the cube and the charge inside the cube.

Calculate the flux linked with a rectangular coil of area 10 cm xx 40 cm , when it is placed in a magnetic field of 0.5 T with its plane (i) normal to the field (ii) parallel to the field (iii) inclined at 30^(@) with the field

A uniform electric field of magnitude E= 100 N C^(-1) exists in the space in x-direction. Calculate the flux of this field through a plane square of edga 10 cm placed in the y-z plane. Take the normal along the positive x-axis to be positive.

At a certain location in the northern hemisphere, the earth's magnetic field has magnitude of 42 mu T and points downwards at 53^(@) to the vertical. Calculate the flux through a horizontal surface of area 2.5 m^(2).[sin 53^(@) = 0.8]