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Considering a Gaussian surface in the form of a sphere at radius r, the electric field has the same magnitude at every point of the sphere and is directed outward. The electric flux is then just the electric field times the area of the sphere.

## What is electric flux through the surface of a sphere due to a point charge line outside it?

Answer: Zero. Explanation: According to gauss law, the net flux passing through a surface is proportional to the charge enclosed within the surface. In this case, the charge is outside the surface so net flux due to this charge is zero.

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

The flux Φ of the electric field →E through any closed surface S (a Gaussian surface) is equal to the net charge enclosed (qenc) divided by the permittivity of free space (ϵ0): Φ=∮S→E⋅ˆndA=qencϵ0.

## What is the electric flux through the sphere due to this charge?

This question already has an answer here:

According to Gauss’s law, the net flux through the sphere is zero because it contains no charge.

## What is the flux of a point charge?

The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field.

## What is flux in sphere?

Considering a Gaussian surface in the form of a sphere at radius r > R , the electric field has the same magnitude at every point of the surface and is directed outward. The electric flux is then just the electric field times the area of the spherical surface.

## Why electric flux through a sphere is zero?

The flux due to the field lines entering is cancelled out by that of the field lines leaving. (because they have opposite signs.) This is why the flux due to external charges is zero.

## What is the flux through the surface?

In physics, specifically electromagnetism, the magnetic flux through a surface is the surface integral of the normal component of the magnetic field B over that surface. It is usually denoted Φ or Φ_{B}. The SI unit of magnetic flux is the weber (Wb; in derived units, volt–seconds), and the CGS unit is the maxwell.

## How do you find the result of electric flux through the sphere?

This makes E and n parallel to each other at each point on the sphere. Thus θ=0 and cosθ = 1 at any point on the surface of the sphere. (c) With E being constant in magnitude, and cosθ = 1 for all points, the total electric flux becomes: Φ = EAcosθ = (1800 N/C) [(4)(3.

## What is the net electric flux through a closed surface?

The net electric flux is zero through any closed surface surrounding a zero net charge. Therefore, if we know the net flux across a closed surface, then we know the net charge enclosed.

## What is the electric flux through a closed surface enclosing an electric dipole?

“The net electric flux through any hypothetical closed surface is equal to 1ε0 times the net charge within that closed surface.” So, the net charge in the above question is zero. Electron dipole is a positive and negative charge of equal magnitude at a specific distance.

## What is the electric flux through the upper face of the box?

The electric flux through the top face (FGHK) is positive, because the electric field and the normal are in the same direction. The electric flux through the other faces is zero, since the electric field is perpendicular to the normal vectors of those faces.

## What is flux through a closed sphere having q charge at centre?

CASE 1: Consider an enclosed spherical surface with a charge q at its centre. From Gauss’ law we can say that the flux through this sphere is q/ϵ0.

## How do you find flux at a point?

If your field does change, then you need to pick a point in time to measure the flux. Measurement: Flux is a total, and is not “per unit area” or “per unit volume”. Flux is the total force you feel, the total number of bananas you see flying by your surface. Think of flux like weight.

## What is the electric flux through the disk?

In the given case the solid angle subtended by the cone subtended by the disc at the point charge is Ω=2π(1−cosθ). So the flux of q which is passing through the surface of the disc is, ϕ=qε0Ω4π=q2ε0(1−cosθ).