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Question: The region between two concentric circle spheres of radius and , respectively (in figure), has volum...

The region between two concentric circle spheres of radius and , respectively (in figure), has volume charge density ρ=Ar\rho = \dfrac{A}{r} , where AA is a constant and rr is the distance from the centre. At the centre of the sphere is a point charge QQ . The value of AA such that the electric field in the region between the spheres will be constant, is:

(A) Q2πa2\dfrac{Q}{{2\pi {a^2}}}
(B) Q2π(b2a2)\dfrac{Q}{{2\pi \left( {{b^2} - {a^2}} \right)}}
(C) Q2π(a2b2)\dfrac{Q}{{2\pi \left( {{a^2} - {b^2}} \right)}}
(D) 2Qπa2\dfrac{{2Q}}{{\pi {a^2}}}

Explanation

Solution

to solve this problem we should know about the Gauss’s theorem and force exerted by electric charge.
Electrostatic force:
F=kq1q2r2F = \dfrac{{k{q_1}{q_2}}}{{{r^2}}} , here kk is constant.
Gauss’s theorem: it states that the net flux through any hypothetical closed surface will equal to 1ε0\dfrac{1}{{{\varepsilon _0}}} times the total charge inside the closed surface.
E.ds=1ε0q\oint {E.ds = \dfrac{1}{{{\varepsilon _0}}}q}

Complete step by step solution:
Let’s assume that a sphere of radius rr and it has thickness drdr lying in the region between two concentric spheres.
From Gauss's theorem, charge inside alone will contribute to the electric field. We get,
So, kQa2=k[Q+ab4πr2drAr]b2\dfrac{{kQ}}{{{a^2}}} = \dfrac{{k\left[ {Q + \int\limits_a^b {4\pi {r^2}dr\dfrac{A}{r}} } \right]}}{{{b^2}}}
Qb2a2=Q+Aab4πrdr\Rightarrow Q\dfrac{{{b^2}}}{{{a^2}}} = Q + A\int\limits_a^b {4\pi rdr}
By integrating the given equation. We get,
Q(b2a21)=2πA(b2a2)\Rightarrow Q\left( {\dfrac{{{b^2}}}{{{a^2}}} - 1} \right) = 2\pi A({b^2} - {a^2})
Q(b2a2a2)=2πA(b2a2)\Rightarrow Q\left( {\dfrac{{{b^2} - {a^2}}}{{{a^2}}}} \right) = 2\pi A({b^2} - {a^2})
A=Q2πa2\Rightarrow A = \dfrac{Q}{{2\pi {a^2}}}
So, we can conclude from the above solution, option (a) is the right answer.

Note:
Gauss’s law is used to solve complex problems. As we have to only find the charge inside the given surface then we can easily calculate the electric field due to these charges. We can find complex problems like electric fields due to infinite long charge carrying wire, electric field due to charge plate, electric field due to charged cylinder. Electric field is a force experienced by a charged particle in the periphery of another charged particle.