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The flux through a cube with a charge inside is the total electric field passing through the cube's surface. It is a measure of the electric field strength and direction at the surface of the cube due to the charge inside.

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6mo ago

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A charge Q is placed at the centre of the cube find flux of the electric field through the six surfaces of the cube?

acc. to gauses law..flux is equal to total charge/e or Q/e as there are six sides of the cube so flux through one surface will be Q/6e


What is the net electric flux through the cube?

The net electric flux through the cube is the total amount of electric field passing through the surface of the cube.


What is the electric flux through one face of the cube?

The electric flux through one face of the cube is the total electric field passing through that face.


What happens when the volume of the sphere is doubled to the total flux through a spherical gaussian surface surrounding a point charge?

You then have an imaginary sphere with twice the volume of the one you imagined only moments earlier. If the point charge is anywhere inside the smaller sphere, and the smaller sphere is anywhere inside the larger one, then the total flux is the same through each of them, although the average flux density through the larger one is less than the average flux density through the smaller one.


What happens to the total flux through a spherical Gaussian surface surrounding a point charge when the charge is moved to another location inside the surface?

The total flux through any closed surface remains constant as long as the totalamount of charge inside it remains constant, regardless of how the surface isshaped, where inside the charge is located or how it's distributed inside.One might even go so far as to say that [ div E = Q ]


Is flux is directly proportional to charge?

Yes, according to Gauss's law, the flux through a closed surface is directly proportional to the charge enclosed by that surface. This is known as the electric flux theorem.


What is the net electric flux through the torus (i.e. doughnut shape) of the figure (Figure 1) Assume that Q100nC and q and minus6.0nC.?

To determine the net electric flux through the torus, we can use Gauss's Law, which states that the electric flux through a closed surface is proportional to the enclosed charge. If the torus does not enclose any charge (meaning the total charge inside is zero), then the net electric flux through the torus will also be zero, regardless of the charges outside it. Given that the charges are ( +100 , \text{nC} ) and ( -6.0 , \text{nC} ), the net charge inside the torus would be ( 100 , \text{nC} - 6.0 , \text{nC} = 94 , \text{nC} ). Therefore, the net electric flux through the torus would be ( \frac{94 , \text{nC}}{\varepsilon_0} ), where ( \varepsilon_0 ) is the permittivity of free space.


What will be value of flux if the surface does not enclose any charge?

If the surface does not enclose any charge, the electric flux through the surface will be zero. This is because electric flux is a measure of the total electric field passing through a surface, and if there are no charges within the surface, there will be no electric field passing through it.


What is the formula for calculating the electric flux through a surface due to a point charge, taking into account the permittivity of free space, represented by the symbol epsilon naught?

The formula for calculating the electric flux through a surface due to a point charge is given by q / , where is the electric flux, q is the charge, and is the permittivity of free space.


How do you define gauss law-?

Gauss's Law states that the total electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. In simpler terms, it describes how the total electric field passing through a closed surface is related to the total charge inside that surface.


What happens to the amount of flux of a single point charge is enclosed with a volume that is increased twice its original values?

The flux through a closed surface enclosing a point charge is independent of the size of the volume it encloses. This is because the electric field due to a point charge follows an inverse square law, resulting in the flux being constant regardless of the size of the enclosing surface.


S1 and S2 are two hollow concentric sphere enclosing charges Q1 and 2Q How will the electric flux through the sphere S1 change if a medium of dielectric?

When a dielectric medium is introduced between the two concentric spheres S1 and S2, the electric flux through S1 will remain unchanged. This is because the electric flux is determined by the charge enclosed within the surface, as described by Gauss's law. Since S1 encloses only charge Q1, the electric flux through S1 is solely dependent on that charge, irrespective of the presence of the dielectric medium. The dielectric affects the electric field within the medium but does not alter the total enclosed charge.