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Module II. Electrostatics
Topic 8. 

Gauss’s Law
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Module II. Electrostatics

Topic 8. Gauss’s Law

Electric flux: is The electric field lines passing through a surface area

Particle Dynamics in the Presence of Electric Fields

In that case the electric flux will be zero. The flux will also be zero if there is no charge inside the box or charge is outside.

what would be the electric flux if you place inside an equal magnitude charge with an opposite sign?

Electric Flux

The electric flux would increase.

What would happen if the particle inside the box increases its magnitude?

ELECTRIC FLUX

Electric flux

The rectangular plate of the given figure has dimensions 0.4m and 0.6m. An electric field of 75 N/C passes through the plate with a deviation of 20° with respect to the plate’s surface. Determine the electric flux magnitude that passes through the rectangular plate.

The square surface shown in the given figure measures 3 mm on each side. It is immersed in a uniform electric field with magnitude E = 2000N/C. The field lines make an angle of 30° with a normal to the surface, as shown. Take that normal to be directed "outward," as though the surface is one face of a box. Calculate the electric flux through the surface.

permittivity constant

If more than one charge is enclosed, you will add them algebraically

is known as the integral form of Gauss’ law

“The total electric flux through any closed surface is equal to the total (net) electric charge enclosed by the surface, divided by ” (Sears & Zemansky, 2013).

Gauss’ Law

The same number of field lines pass through both of these area elements

“The Gaussian surface is known as a closed surface in three-dimensional space such that the flux of a vector field is calculated, This imaginary sphere can also be an irregular shape".

Gaussian Surface

Where L is length, A is the area and V is the volume of the object.

When the charge is distributed along an object, the charge density needs to be considered

Volume charge Density

Surface charge Density

Linear charge Density

Gauss´ LawApplications

r IS THE DISTANCE OF THE POINT

r IS THE DISTANCE OF THE POINT

Electric field for different charge distributions

Gauss Law

Obtain the electric field magnitude at any point of an infinite charged plate with uniform charge per unit area of

A sphere 0.04 m of radius has a charge of 6x10-9 placed on its Surface, what is the Electric field intensity at a) the Surface, b) 0.015m outside the Surface and c) 0.015m inside the surface using Gauss´s Law.

An infinite line charge produces a field of 9 × 10 4N/C at a distance of 2 cm. Calculate the linear charge density and find the total charge enclosed.

The charges in the following image have a magnitude of +3nC (red charges) and -4nC (blue charges). Calculate the electric flux through each of the gaussian surfaces represented by the circles of different colors.

The charge density on each of two parallel plates is of 5x10^-6 C/m^2. Find the intensity of the electric field between both plates.

A solid insulating sphere has a radius of 5 cm, the charge is uniformly distributed and has a magnitude of 6 nC. Calculate the electric field at a) 3.4 cm from the center and b) 6.8 cm from the center.

¡THANKS!

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