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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.

## 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 cm ^{2} = 4 x 10^{–}^{2} m^{2}.**