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"Gauss's law in electrostatics is true for any closed surface no matter what its shape of size is." Justify the statement with the help of a suitable example.?
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"Gauss's law in electrostatics is true for any closed surface no matte...
Gauss's Law :
The total electric  flux over a closed surface in vacuum is equal to 1/ε₀ the total charge enclosed by the surface.
if q is the charge enclosed by surface S, The total electric flux through the surface S is Φ=q/ε₀
while measuring the electric flux, surface area is more important than its volume or  its size.
Example: 
let us consider a closed surface S encloses a point charges +q1,-q2,+q3 and -q4, the total charge enclosed is q= q1-q2+q3-q4
∴electric flux across S will be Φ=q/ε₀=(q1-q2+q3-q4)/ε₀
The total electric flux is zero if no charge is enclosed by the surface.
conclusion:
The closed surface chosen for application of Gauss's law is called as Gaussian surface. It can be of any shape or size.


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"Gauss's law in electrostatics is true for any closed surface no matte...
Gauss's Law in Electrostatics: True for any closed surface

Introduction:
Gauss's law in electrostatics is one of the fundamental principles in the study of electric fields. It relates the electric flux passing through a closed surface to the total charge enclosed by that surface. The statement "Gauss's law in electrostatics is true for any closed surface no matter what its shape or size is" can be justified through a suitable example.

Example:
Consider a point charge q located at the center of a sphere of radius r. We want to determine the electric field at any point on the surface of the sphere using Gauss's law.

Step 1: Choose a Gaussian surface
To apply Gauss's law, we choose a closed surface that encloses the point charge. In this case, we choose a spherical surface with the point charge at its center.

Step 2: Calculate the electric flux
The electric flux passing through the chosen Gaussian surface is given by the product of the electric field and the surface area. Since the electric field is radial and constant over a spherical surface, the electric field is directed outward and perpendicular to the surface at every point. Therefore, the electric flux passing through the entire surface of the sphere is given by Φ = E * 4πr².

Step 3: Determine the charge enclosed
The charge enclosed by the Gaussian surface is simply the charge q of the point charge located at the center of the sphere.

Step 4: Apply Gauss's law
According to Gauss's law, the electric flux passing through a closed surface is equal to the total charge enclosed divided by the permittivity of free space (ε₀). Therefore, we can write Φ = q / ε₀.

Step 5: Equate the expressions
Equating the expressions for electric flux obtained from the calculation and Gauss's law, we have E * 4πr² = q / ε₀.

Step 6: Solve for the electric field
Rearranging the equation, we find the electric field E = q / (4πε₀r²). This is the electric field at any point on the surface of the sphere.

Conclusion:
The above example demonstrates the application of Gauss's law in electrostatics to determine the electric field on the surface of a sphere. The result obtained is independent of the shape or size of the Gaussian surface chosen. This illustrates the generality of Gauss's law, which holds true for any closed surface. Whether the surface is a sphere, cube, or any other shape, as long as it encloses the charge or charges of interest, Gauss's law can be applied to relate the electric flux passing through the surface to the total charge enclosed.
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"Gauss's law in electrostatics is true for any closed surface no matter what its shape of size is." Justify the statement with the help of a suitable example.?
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