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Reason: According to Gauss theorem, total electric flux through a closed surface enclosing a charge is equal to 1/ε0 times the magnitude of the charge enclosed. (A): A point charge is lying at the centre of a cube of each side. The electric flux emanating from each surface of the cube is 1th6 total flux.

## What is the electric flux through one face of a cube containing a charge Q at the cube’s center?

So, we arrive at the conclusion that, for the original problem geometry, where a charge q was placed at one corner of the cube, the net flux through each of the faces marked by 1,2 and 3, is F/3=q/24episilon0.

## What is the electric flux through the surface of the cube?

Electric flux through the closed surface ise1 times the charge enclosed by it. Where, e is electrical permittivity. Here, Electric flux=6eq.

## What is the electric flux associated with one of the faces of cube?

Therefore, the electric flux associated with one of the faces of the cube is q6ε0 .

## What is the charge enclosed by the cube?

If the frame of reference is at the centre of the cube, the total flux passing through the cube would be zero (since both the faces P and Q would have equal distances from the reference and would cancel the electric flux) and hence the charge enclosed by the cube would be zero.

## What is the electric flux through a cubicle Gaussian surface enclosed on an electric dipole?

Answer: The net total electric flux will be zero. Hence net electric flux through all surfaces of the cube is zero.

## When was electric flux associated?

The negative flux just equals in magnitude the positive flux, so that the net, or total, electric flux is zero. If a net charge is contained inside a closed surface, the total flux through the surface is proportional to the enclosed charge, positive if it is positive, negative if it is negative.

## When the electric flux associated with closed surface becomes positive zero or negative?

When the electric flux associated with closed surface becomes positive, zero or negative ? Where θ is angle between →E and →S. ∴ Flux associated with surface is zero. (ii) If θ<90∘, then cosθ>0 (negative) hence flux is positive.

## What is the electric flux through any one side of the cube if a point particle with charge q is placed inside the cube but not at its center?

Since the electric field is perpendicular to the opening the angle between the electric field and any area vector in the round surface will be 90° and hence there will be no flux through the sides.

## What is electric flux through a surface?

In electromagnetism, electric flux is the measure of the electric field through a given surface, although an electric field in itself cannot flow. It is a way of describing the electric field strength at any distance from the charge causing the field.