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Electrostatics question

2012 · Shift 0 · Q60
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Electrostatics question

2012 · Shift 0 · Q60

JEE MainPhysicsElectrostaticsMCQ+4 / −1
In a uniformly charged sphere of total charge QQQ and radius R,R,R, the electric field EEE is plotted as function of distance from the center. The graph which would correspond to the above will be:
  1. A
    AIEEE 2012 Physics - Electrostatics Question 225 English Option 1
  2. B
    AIEEE 2012 Physics - Electrostatics Question 225 English Option 2
  3. C
    AIEEE 2012 Physics - Electrostatics Question 225 English Option 3
  4. D
    AIEEE 2012 Physics - Electrostatics Question 225 English Option 4
View written solutionFree

Correct answer: C

  1. Electric field inside a uniformly charged solid sphere

For a uniformly charged sphere of radius RRR and total charge QQQ, the volume charge density is

ρ=Q43πR3.\rho = \frac{Q}{\frac{4}{3}\pi R^3}.ρ=34​πR3Q​.

Using Gauss’s law for a point at distance r<Rr<Rr<R from the center:

E(4πr2)=Qenclε0E(4\pi r^2)=\frac{Q_{\text{encl}}}{\varepsilon_0}E(4πr2)=ε0​Qencl​​

where

Qencl=ρ⋅43πr3=Qr3R3.Q_{\text{encl}}=\rho \cdot \frac{4}{3}\pi r^3 = Q\frac{r^3}{R^3}.Qencl​=ρ⋅34​πr3=QR3r3​.

So,

E=14πε0QrR3,(r<R)E = \frac{1}{4\pi\varepsilon_0}\frac{Qr}{R^3}, \qquad (r<R)E=4πε0​1​R3Qr​,(r<R)

Hence, inside the sphere, EEE increases linearly with rrr.

  1. Electric field outside the sphere

For r>Rr>Rr>R, the sphere behaves like a point charge QQQ at the center:

E=14πε0Qr2,(r>R)E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}, \qquad (r>R)E=4πε0​1​r2Q​,(r>R)

So, outside the sphere, EEE decreases as 1r2\dfrac{1}{r^2}r21​.

  1. Value at the surface

At r=Rr=Rr=R,

E(R)=14πε0QR2.E(R)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}.E(R)=4πε0​1​R2Q​.

The field is continuous at the surface.

  1. Shape of the graph

Therefore, the correct EEE vs rrr graph must:

  • start from E=0E=0E=0 at r=0r=0r=0,
  • increase linearly up to r=Rr=Rr=R,
  • attain maximum at r=Rr=Rr=R,
  • then decrease as a curved 1/r21/r^21/r2 graph for r>Rr>Rr>R.

So the correct option is the graph showing a straight line from origin to r=Rr=Rr=R, followed by a decreasing curve beyond RRR.

Thus, the correct answer is C.

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