Answer
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Hint: We need to find the relation between the electric field due to a charge, the number of electric field lines, and the dielectric constant in the system of a charge. We can relate this by means of the formula for electric field strength from Coulomb’s law.
Complete step-by-step solution:
The number of electric field lines evolving from a charge is the electric flux of the system. It is defined as the number of electric field lines produced by a charge enclosed in a given area surrounded by the charge at a distance say $r$ from the charge.
We know that the electric field strength is given from the coulomb’s law as the force experienced per unit charge due to another charge at a distance $r$ . It is given as;
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}$
Now, we can find the number of field lines due to the charge $q$ in an area enclosed within $r$ units from the charge.
![](https://www.vedantu.com/question-sets/e4dffb3c-ec01-427b-9446-9aa9e49016ae464994275256561778.png)
The electric flux is given as-
$A = 4\pi {r^2}$
Now, we can find the number of field lines or electric flux for a charge $0.5\;C$ placed in a dielectric medium with $K = 10$, is given as,
$E = \dfrac{1}{{4\pi \varepsilon }}\dfrac{q}{{{r^2}}}$
$ \Rightarrow E = \dfrac{k}{K}\dfrac{q}{{{r^2}}}$
where, $k = \dfrac{1}{{4\pi {\varepsilon _0}}}$ and $K = \dfrac{\varepsilon }{{{\varepsilon _0}}}$
Also, $A = 4\pi {r^2}$
Now,
$\phi = E.A$
$ \Rightarrow \phi = \dfrac{k}{K}\dfrac{q}{{{r^2}}}4\pi {r^2}$
Upon substituting the values we get,
$\phi = \dfrac{{9 \times {{10}^9} \times 0.5 \times 4\pi }}{{10}}$
$ \Rightarrow \phi = 5.65 \times {10^9}$
The electric flux or the number of field lines passing through the surface of the sphere at a distance $r$ from the charge is $\phi = 5.65 \times {10^9}$ .
The correct answer is option (A).
Note: The idea of electric flux density is more generally used than the number of field lines or the electric flux. It is the measure of the concentration of field lines at distance from the charge. For a parallel field, the electric flux is not dependent on the distance from the source.
Complete step-by-step solution:
The number of electric field lines evolving from a charge is the electric flux of the system. It is defined as the number of electric field lines produced by a charge enclosed in a given area surrounded by the charge at a distance say $r$ from the charge.
We know that the electric field strength is given from the coulomb’s law as the force experienced per unit charge due to another charge at a distance $r$ . It is given as;
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}$
Now, we can find the number of field lines due to the charge $q$ in an area enclosed within $r$ units from the charge.
![](https://www.vedantu.com/question-sets/e4dffb3c-ec01-427b-9446-9aa9e49016ae464994275256561778.png)
The electric flux is given as-
$A = 4\pi {r^2}$
Now, we can find the number of field lines or electric flux for a charge $0.5\;C$ placed in a dielectric medium with $K = 10$, is given as,
$E = \dfrac{1}{{4\pi \varepsilon }}\dfrac{q}{{{r^2}}}$
$ \Rightarrow E = \dfrac{k}{K}\dfrac{q}{{{r^2}}}$
where, $k = \dfrac{1}{{4\pi {\varepsilon _0}}}$ and $K = \dfrac{\varepsilon }{{{\varepsilon _0}}}$
Also, $A = 4\pi {r^2}$
Now,
$\phi = E.A$
$ \Rightarrow \phi = \dfrac{k}{K}\dfrac{q}{{{r^2}}}4\pi {r^2}$
Upon substituting the values we get,
$\phi = \dfrac{{9 \times {{10}^9} \times 0.5 \times 4\pi }}{{10}}$
$ \Rightarrow \phi = 5.65 \times {10^9}$
The electric flux or the number of field lines passing through the surface of the sphere at a distance $r$ from the charge is $\phi = 5.65 \times {10^9}$ .
The correct answer is option (A).
Note: The idea of electric flux density is more generally used than the number of field lines or the electric flux. It is the measure of the concentration of field lines at distance from the charge. For a parallel field, the electric flux is not dependent on the distance from the source.
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