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JEE Main 20238 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
8 April
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2023 · 8 April · Morning Shift · Physics · Electrostatics

Graphical variation of electric field due to a uniformly charged...

Graphical variation of electric field due to a uniformly charged insulating solid sphere of radius R,R,with distance rr from the centre OO is represented by:

Graphical variation of electric field due to a uniformly charged insulating solid sphere of radius R, with distance r

Options

  1. AThe first and second ionization constants of a weak dibasic acid H_2 A are 8.1 × 10^ -8 and 1.0 × 10^ -13 respectively.
  2. BThe first and second ionization constants of a weak dibasic acid H_2 A are 8.1 × 10^ -8 and 1.0 × 10^ -13 respectively.
  3. CThe first and second ionization constants of a weak dibasic acid H_2 A are 8.1 × 10^ -8 and 1.0 × 10^ -13 respectively.
  4. DThe first and second ionization constants of a weak dibasic acid H_2 A are 8.1 × 10^ -8 and 1.0 × 10^ -13 respectively.

Correct answer

A. The first and second ionization constants of a weak dibasic acid H_2 A are 8.1 × 10^ -8 and 1.0 × 10^ -13 respectively.

Step-by-step solution

The electric field inside a solid insulated sphere is given by&

Einside=ρr3ε0{E}_{inside} = \dfrac{ρr}{3{ε}_{0}} for r<Rr < R where ρρ is& volume& charge density.

Hence,& the electric field varies linearly with the distance as

Einside∝r1{E}_{inside} ∝ r^{1}

Now, the electric field for r>Rr > R is given by&

Eoutside=Q4πε0r2{E}_{outside} = \dfrac{Q}{4\pi{ε}_{0}r^{2}}

Eoutside∝1r2{E}_{outside} ∝ \dfrac{1}{r2}

Thus, the graph starts decreasing as a power law graph as shown in the figure for the region r>Rr > R.

Hence, this graph is the right option.

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