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Question: The number of electric field lines leaving the positive charge \(0.5\;C\) placed in the medium of di...

The number of electric field lines leaving the positive charge 0.5  C0.5\;C placed in the medium of dielectric constant K=10K = 10 are:
(A) 5.65×1095.65 \times {10^9}
(B) 1.13×10111.13 \times {10^{11}}
(C) 9×1099 \times {10^9}
(D) 8.85×10128.85 \times {10^{ - 12}}

Explanation

Solution

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 rr 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 rr . It is given as;
E=14πε0qr2E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}
Now, we can find the number of field lines due to the charge qq in an area enclosed within rr units from the charge.

The electric flux is given as-
A=4πr2A = 4\pi {r^2}
Now, we can find the number of field lines or electric flux for a charge 0.5  C0.5\;C placed in a dielectric medium with K=10K = 10, is given as,
E=14πεqr2E = \dfrac{1}{{4\pi \varepsilon }}\dfrac{q}{{{r^2}}}
E=kKqr2\Rightarrow E = \dfrac{k}{K}\dfrac{q}{{{r^2}}}
where, k=14πε0k = \dfrac{1}{{4\pi {\varepsilon _0}}} and K=εε0K = \dfrac{\varepsilon }{{{\varepsilon _0}}}
Also, A=4πr2A = 4\pi {r^2}
Now,
ϕ=E.A\phi = E.A
ϕ=kKqr24πr2\Rightarrow \phi = \dfrac{k}{K}\dfrac{q}{{{r^2}}}4\pi {r^2}
Upon substituting the values we get,
ϕ=9×109×0.5×4π10\phi = \dfrac{{9 \times {{10}^9} \times 0.5 \times 4\pi }}{{10}}
ϕ=5.65×109\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 rr from the charge is ϕ=5.65×109\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.