# What is the electric flux through a cube of side 5 cm which encloses an electric dipole?

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## What is the electric flux through a cube of side 1 cm which encloses an electric dipole charges PC?

Since net charge enclosed in the surfaces bound by the cube is zero(dipole consists of equal and opposite charges), according to the Gauss Law, it can be claimed that the net flux through the cube is zero.

## What is the electric flux through the side A of the cube?

The flux through the surface of this cube is just q/ϵ0 by Gauss’s Law since it is a closed surface containing the charge.

## What is the electric flux through a cylinder that encloses an electric dipole?

Answer: You can say that the electric flux will be zero as result of Gauss’s Law. Enclosing any number of charges who’s sum equals zero will result in a surface flux of zero.

## What do you mean by electric dipole moment?

qd is defined as the electric dipole moment. Its magnitude indicates the maximum torque exerted upon a given electric dipole per unit value of the surrounding electric field in a vacuum. The electric dipole moment, a vector, is directed along the line from negative charge toward positive charge.

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## How does electric flux due to a point charge?

How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased? … That is, on increasing the radius of the gaussian surface, charge q remains unchanged. So, flux through the gaussian surface will not be affected when its radius is increased.

## What are the points at which electric potential of a dipole has a minimum value?

When the angle between the dipole moment and electric field is zero then the potential energy of electric dipole is minimum.

## What is the flux through a cube of side A If a point charge?

Complete step by step solution:

According to gauss law, the electric flux through a closed surface is equal to [dfrac{q}{{{varepsilon _circ }}}], if q is the charge enclosed in it. If the charge ‘q ‘is placed at one of the corners of the cube, it will be divided into 8 such cubes.

## What is the net flux through the cube of side 20 cm?

What is the net flux of the uniform electric field of question 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes? The area of each face out of the six faces of the cube = 20 x 20 = 400 cm2 = 4 x 102 m2.