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Electric Flux Through Cone Or Disc - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
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Electric flux through cone or disc is considered one the most difficult concept.
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25 Questions around this concept.
Solve by difficulty
A cone of base radius R and height $h$ is located in a uniform electric field $E$ parallel to its base. The electric flux entering the cone is :
Electric charge is uniformly distributed along a long straight wire of radius 1 mm. The charge per cm length of the wire is Q coulomb. Another cylindrical surface of radius 50 cm and length 1 m symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is
Shown below is a distribution of charges. The flux of the electric field due to these charges through the surface S is
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If the electric flux entering and leaving an enclosed surface respectively is and
, the electric charge inside the surface will be:
Let be the charge density distribution for a solid sphere of radius
and total charge
. For a point
inside the sphere at a distance
from the centre of the sphere, the magnitude of the electric field is
Let there be a spherically symmetric charge distribution with charge density varying as where
is the distance from the origin. The electric field at a distance
from the origin is given by
$q_1, q_2, q_3, q_4$ are point charges located at points as shown in the figure and s is a spherical Gaussian surface of radius $R$. Which of the following is true according to Gauss's law
Concepts Covered - 1
Electric flux through cone or disc
Electric flux through cone or disc-
There are several case for electric flux calculation. In this concept we will discuss one very important and complex case which is ''Electric flux through cone or disc''. For this let us consider a point charge at a distance 'a' from a disc of radius R as shown in the given figure.
Let us consider an elemental ring of radius "y" and width "dy". Area of this ring(strip) is $d A=2 \pi y d y$.
Electric field due to $q$ at this elemental ring,
$
E=\frac{1}{4 \pi \varepsilon_0} \frac{q}{\left(a^2+y^2\right)}
$
If $d \phi$ is the flux passing through this elemental ring, we have
$
\begin{aligned}
d \phi & =E d A \cos \theta \\
& =\left(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{\left(a^2+y^2\right)}\right)(2 \pi y d y)\left(\frac{a}{\left(a^2+y^2\right)^{1 / 2}}\right) \\
& =\frac{q a}{2 \varepsilon_0} \cdot\left(\frac{y d y}{\left(a^2+y^2\right)^{3 / 2}}\right)
\end{aligned}
$
To obtain total flux, we should integrate this expression over the whole area of the ring, So the total flux is can be given as -
$
\phi=\int d \phi=\frac{q a}{2 \varepsilon_0} \int_0^R \frac{y d y}{\left(a^2+y^2\right)^{3 / 2}}
$
On integration we get,
$
\phi=\frac{q}{2 \varepsilon_0}\left(1-\frac{a}{\sqrt{a^2+R^2}}\right)
$
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