A cube of side a is placed such that the...

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

Similar Questions

Explore conceptually related problems

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 flux phi_(E) through 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 -

Due to a charge inside a cube the electric field is E_(x)=600 x^(1//2), E_(y)=0, E_(z)=0 . The charge inside the cube is (approximately):

If the electric field components due to electric charge in cube shown in figure are E_x = 600x^(1/2) and E_y = 0 and E_z = 0 , then charge within the cube…………

The electric field components in Fig. are E_(x)=ax^(1//2), E_(y)=E_(z)=0 , in which alpha=800N//C m^(1//2) . Calculate the flux through the cube

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 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

Similar Questions

Recommended Questions

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