Define electric flux and write its SI unit . The electric field components in the figure shown are : `E_(x)=alphax,E_(y)=0,E_(z)=0` where `alpha=(100N)/(Cm)`.
Calculate the charge within the cube , assuming a = 0.1 m .

The electric components in the figure are E_(x)=alphax^(1//2) ,E_(y)=0,E_(z)=0 where alpha=800N//m^(2) if a=0.1m is the side of cube then the charge with in the cube is

(a) Define electric flux. Write its SI units. (b) The electric field components due to a charge inside the cube of side 0.1 m are as shown : E_(x)=alpha x, where alpha=500N//C-m E_(y)=0, E_(z)=0 . Calculate (i) the flux through the cube, and (ii) the charge inside the cube.

Using Gauss’s law, derive expression for intensity of electric field at any point near the infinitely long straight uniformly charged wire. The electric field components in the following figure are E_(x) = alphax, E_(y) = 0, E_(z) = 0, " in which " alpha = 400 N//C m. Calculate (i) the electric flux through the cube, and (ii) the charge within the cube assume that a = 0.1m.

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.

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.

Net electric flux through a sphere of radius 1 m placed in a uniform electric field as shown in the figure is → E = 103 N/C

A cube of side a is placed such that the nearest face , which is parallel to the yz plane , is at a distance a from the origin . The electric field components are E_(x) = alpha x ^(1//2) , E_(y) = E_(z) = 0 . The charge within the cube is

Due to a charge inside the cube, the electric field is: E_(x) = 600x, E_(y) = 0, E_(z) = 0. The charge inside the cube is nearly -