CameraIcon

SearchIcon

MyQuestionIcon

Ampere's Circuital Law

the electric ...

Question

the electric field components in figure below are Ex=800x^{2^{}} N/c E_y=Ez=0 calculate a) the flux througu cube b) the charge within the cube

Open in App

Solution

Suggest Corrections

thumbs-up

7

mid-banner-image

mid-banner-image

Similar questions

Q. The electric field components in figure are

E_{x} = b \sqrt{x}, E_{y} = E_{z} = 0,

Calculate the net flux through the cube.

Q. The electric field components in the given figure are

E_{x} = α x^{1 / 2}, E_{y} = E_{z} = 0

α = 800 N C^{- 1} m^{- 1 / 2}

. The charge within the cube is, if net flux through the cube is

1.05 N m^{2} C^{- 1}

a = 0.1 m

)

The electric field components in the figure are $E_{x} = α x^{1 / 2}, E_{y} = 0, E_{z} = 0$ where $α = 800 N / m^{2}$ . If a = 0.1 m is the side of cube then the charge within the cube is:

Q. The electric field components due to a charge inside the cube of side 0.1 m are shown in figure.
where,

E_{x} = α

α

E_{y}

E_{z} = 0

(a) the flux through the cube and
(b) the charge inside the cube.

The electric field components in the given figure are $E_{x} = α x^{\frac{1}{2}}, E_{y} = E_{z} = 0$ in which $α = 800 N C^{- 1} m^{\frac{- 1}{2}}$ . The charge within the cube is if net flux through the cube is $1.05 N m^{2} C^{- 1}$ (assume a = 0.1 m)

View More

Join BYJU'S Learning Program

Select...

similar_icon

Related Videos

lock

Ampere's Law

PHYSICS

Watch in App

explore_more_icon

Explore more

Ampere's Circuital Law

Standard XII Physics

Join BYJU'S Learning Program

Select...