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Electrostatics question

2019 · 12 Apr · Shift 2 · Q55
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Electrostatics question

2019 · 12 Apr · Shift 2 · Q55

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by ρ\rhoρ(r) = kr, where r is the distance from the centre. Two charges A and B, of –Q each, are placed on diametrically opposite points, at equal distance, aaa from the centre. If A and B do not experience any force, then :
  1. A
    a=8−1/4Ra = {8^{ - 1/4}}Ra=8−1/4R
  2. B
    a=2−1/4Ra = {2^{ - 1/4}}Ra=2−1/4R
  3. C
    a=3R21/4a = {{3R} \over {{2^{1/4}}}}a=21/43R​
  4. D
    a=R3a = {R \over {\sqrt 3 }}a=3​R​
View written solutionFree

Correct answer: A

  1. Given charge density and total charge

The sphere has radius RRR and volume charge density

ρ(r)=kr.\rho(r)=kr.ρ(r)=kr.

The total charge inside the sphere is given as 2Q2Q2Q.

Using spherical shells,

dq=ρ(r) dV=kr⋅4πr2dr=4πkr3dr.dq=\rho(r)\, dV = kr\cdot 4\pi r^2 dr = 4\pi k r^3 dr.dq=ρ(r)dV=kr⋅4πr2dr=4πkr3dr.

So,

2Q=∫0R4πkr3 dr=4πk[r44]0R=πkR4.2Q=\int_0^R 4\pi k r^3\,dr =4\pi k\left[\frac{r^4}{4}\right]_0^R =\pi k R^4.2Q=∫0R​4πkr3dr=4πk[4r4​]0R​=πkR4.

Hence,

k=2QπR4.k=\frac{2Q}{\pi R^4}.k=πR42Q​.
  1. Charge enclosed within radius aaa

A charge −Q-Q−Q is placed at distance aaa from the centre. Since the distribution is spherically symmetric, only the charge enclosed within radius aaa contributes to the electric field at that point.

Thus,

qenc(a)=∫0a4πkr3 dr=πka4.q_{\text{enc}}(a)=\int_0^a 4\pi k r^3\,dr =\pi k a^4.qenc​(a)=∫0a​4πkr3dr=πka4.

Substitute kkk:

qenc(a)=π⋅2QπR4a4=2Qa4R4.q_{\text{enc}}(a)=\pi \cdot \frac{2Q}{\pi R^4} a^4 =2Q\frac{a^4}{R^4}.qenc​(a)=π⋅πR42Q​a4=2QR4a4​.
  1. Electric field at A due to the charged sphere

By Gauss's law,

Esphere(a)=14πε0qenc(a)a2=14πε02Qa4/R4a2=14πε02Qa2R4.E_{\text{sphere}}(a)=\frac{1}{4\pi\varepsilon_0}\frac{q_{\text{enc}}(a)}{a^2} =\frac{1}{4\pi\varepsilon_0}\frac{2Q a^4/R^4}{a^2} =\frac{1}{4\pi\varepsilon_0}\frac{2Q a^2}{R^4}.Esphere​(a)=4πε0​1​a2qenc​(a)​=4πε0​1​a22Qa4/R4​=4πε0​1​R42Qa2​.

This field is directed radially outward from the centre. Since charge at A is negative, force on A due to sphere is toward the centre.

Therefore,

F_{\text{sphere on A}}=Q E_{\text{sphere}}= rac{1}{4\pi\varepsilon_0}\frac{2Q^2 a^2}{R^4}.
  1. Force on A due to charge B

The two charges AAA and BBB are at diametrically opposite points, each at distance aaa from the centre. Hence separation between them is

AB=2a.AB=2a.AB=2a.

Both charges are −Q-Q−Q, so they repel each other.

Magnitude of force on A due to B:

FBA=14πε0Q2(2a)2=14πε0Q24a2.F_{BA}=\frac{1}{4\pi\varepsilon_0}\frac{Q^2}{(2a)^2} =\frac{1}{4\pi\varepsilon_0}\frac{Q^2}{4a^2}.FBA​=4πε0​1​(2a)2Q2​=4πε0​1​4a2Q2​.

This force is away from B, i.e. outward along the diameter.

  1. Condition for zero net force

For A (and similarly B) to experience no net force,

Fsphere on A=FBA.F_{\text{sphere on A}}=F_{BA}.Fsphere on A​=FBA​.

So,

14πε02Q2a2R4=14πε0Q24a2.\frac{1}{4\pi\varepsilon_0}\frac{2Q^2 a^2}{R^4} = \frac{1}{4\pi\varepsilon_0}\frac{Q^2}{4a^2}.4πε0​1​R42Q2a2​=4πε0​1​4a2Q2​.

Cancel common factors:

2a2R4=14a2.\frac{2a^2}{R^4}=\frac{1}{4a^2}.R42a2​=4a21​.

Thus,

8a4=R4.8a^4=R^4.8a4=R4.

Hence,

a=R81/4=8−1/4R.a=\frac{R}{8^{1/4}}=8^{-1/4}R.a=81/4R​=8−1/4R.
  1. Matching with options
a=8−1/4Ra=8^{-1/4}Ra=8−1/4R

which corresponds to Option A.

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