1 Crore+ students have signed up on EduRev. Have you? 
Gauss law:
Hence, the correct option is (4)
A point charge Q is located on the axis of a disk of radius R at a distance b from the plane of the disk as shown in below figure. Find the value of R, if 1/4th of the electric flux from the charge passes through the disk?
Concept:
We know that the flux passing through a segment which makes angle θ at the center of the circle is:
Calculation:
Given if 1/4^{th }of the flux passes through the disk
The electric flux from a cube of side ‘a’ is ‘Φ’. What will be its value if the side of the cube is made ‘2a’ and the charge enclosed is made half?
CONCEPT:
Gauss's Law for electric field: It states that the total electric flux emerging out of a closed surface is directly proportional to the charge enclosed by this closed surface. It is expressed as:
ϕ = q/ϵ_{0}
Where ϕ is the electric flux, q is the charge enclosed in the closed surface and ϵ_{0} is the electric constant.
Given that:
Consider a charge 'q' placed inside a cube of side 'a'.
The electric flux according to Gauss's law,
⇒ ϕ = q/ϵ_{0}
If the charge enclosed is halved, then
⇒ q′ = q/2
Therefore, the new electric flux associated with this,
Concept:
ϕ α Q
ϕ = Q / ϵ_{0}
Here,
Q = charge enclosed by the surface
ϵ_{0} = permittivity of free space
From the above explanation, we can see that according to Gauss law
ϕ = Q/ε_{0}
Whereas, electric flux through small area dA is the product of the electric field and the area its passing through (dA)
∴ϕ = ∮E.dA = Q/ε_{0}
Thus the given equation represents Gauss’s Law for electricity
According to gauss theorem, the electric flux on a closed surface depends on?
CONCEPT
Where, Φ = electric flux, Q_{in} = charge enclosed the sphere, ϵ_{0} = permittivity of space (8.85 × 10^{12} C^{2}/Nm^{2}), dS = surface area
According to gauss’s law,
The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ϵ_{0}.
Electric flux on a closed surface only depends on the enclosed charge.
∴ Option 2 is correct
The net electric flux through a spherical surface having 10 μC charge at the centre is
CONCEPT:
ϕ = q/ϵ_{0}
Where ϕ is the electric flux, q is the charge enclosed in the closed surface and ϵ0 is the permittivity of free space
The charge, q = 10 μC = 10 × 10^{6} C
From Gauss's law, electric flux
If a unit positive charge is placed inside a sphere of radius r, then the electric flux through the sphere will be:
CONCEPT:
Gauss's law:
⇒ ϕ = Q/ϵ_{o}
Where ϕ = electric flux linked with a closed surface, Q = total charge enclosed in the surface, and ϵ_{o} = permittivity
Given Q = 1, and r = radius of the sphere
By the Gauss law, if the total charge enclosed in a closed surface is Q, then the total electric flux associated with it will be given as,
⇒ ϕ = Q/ϵ_{o} (1)
By equation 1 the total flux linked with the sphere is given as,
Hence, option 3 is correct.
The total electric flux through a closed surface in which a certain amount of charge is placed depends on the:
CONCEPT:
Gauss's law:
According to Gauss law, the total electric flux linked with a closed surface called Gaussian surface is 1/ϵ_{o} the charge enclosed by the closed surface.
⇒ ϕ = Q/ϵ_{o}
Where ϕ = electric flux linked with a closed surface, Q = total charge enclosed in the surface, and ϵ_{o} = permittivity
Gauss's law:
⇒ ϕ = Qϵ_{o} (1)
The charge enclosed in the two cubes A and B is q. If ratio of the surface area of cubes A to cube B is 1 : 2, then the ratio of the flux associated with cube A to cube B will be:
Concept:
Gauss's law:
According to Gauss law, the total electric flux linked with a closed surface called Gaussian surface is 1/ϵ_{o} the charge enclosed by the closed surface.
⇒ ϕ = Q/ϵ_{o}
Where ϕ = electric flux linked with a closed surface, Q = total charge enclosed in the surface, and ϵ_{o} = permittivity
Calculation:
Given Q_{A} = Q_{B} = q, and A_{A} : A_{B} = 1 : 2
Where A_{A} and A_{B} = surface area of the cube A and the cube B respectively
By the Gauss law, if the total charge enclosed in a closed surface is Q, then the total electric flux associated with it will be given as,
⇒ ϕ = Q/ϵ_{o} (1)
By equation 1 it is clear that the total flux linked with the closed surface in which a certain amount of charge is placed does not depend on the shape and size of the surface.
By equation 1 the total flux associated with cube A is given as,
By equation 1 the total flux associated with cube B is given as,
By equation 2 and equation 3,
Hence, option 3 is correct.
11 videos46 docs62 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
11 videos46 docs62 tests









