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Electrostatics question

2012 · Shift 0 · Q59
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Electrostatics question

2012 · Shift 0 · Q59

JEE MainPhysicsElectrostaticsMCQ+4 / −1
This question has statement- 111 and statement-2.2.2. Of the four choices given after the statements, choose the one that best describe the two statements. An insulating solid sphere of radius RRR has a uniformly positive charge density ρ\rhoρ. As a result of this uniform charge distribution there is a finite value of electric potential at the center of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinite is zero. Statement- 1:1:1: When a charge qqq is take from the centre of the surface of the sphere its potential energy changes by qρ3ε0{{q\rho } \over {3{\varepsilon _0}}}3ε0​qρ​ Statement-2:2:2: The electric field at a distance r(r<R)r\left( {r \lt R} \right)r(r<R) from the center of the sphere is ρr3ε0.{{\rho r} \over {3{\varepsilon _0}}}.3ε0​ρr​.
  1. A
    Statement- 111 is true, Statement-222 is true; Statement-222 is not the correct explanation of Statement-111.
  2. B
    Statement 111 is true, Statement 222 is false.
  3. C
    Statement 111 is false, Statement 222 is true.
  4. D
    Statement- 111 is true, Statement-222 is true; Statement-222 is the correct explanation of Statement-111.
View written solutionFree

Correct answer: C

  1. Given data

An insulating solid sphere of radius RRR has uniform volume charge density ρ\rhoρ.

We need to check:

  • Statement 1: When a charge qqq is taken from the centre to the surface, its potential energy changes by qρ3ε0\dfrac{q\rho}{3\varepsilon_0}3ε0​qρ​.
  • Statement 2: For r<Rr<Rr<R, electric field is E(r)=ρr3ε0.E(r)=\frac{\rho r}{3\varepsilon_0}.E(r)=3ε0​ρr​.
  1. Check Statement 2

Using Gauss's law for a Gaussian sphere of radius r<Rr<Rr<R:

  • Enclosed charge: qenc=ρ⋅43πr3q_{\text{enc}}=\rho \cdot \frac{4}{3}\pi r^3qenc​=ρ⋅34​πr3

  • By symmetry, electric field is radial and constant on the Gaussian surface: E(4πr2)=qencε0E(4\pi r^2)=\frac{q_{\text{enc}}}{\varepsilon_0}E(4πr2)=ε0​qenc​​

So, E(4πr2)=ρ⋅43πr3ε0E(4\pi r^2)=\frac{\rho \cdot \frac{4}{3}\pi r^3}{\varepsilon_0}E(4πr2)=ε0​ρ⋅34​πr3​

Hence, E=ρr3ε0E=\frac{\rho r}{3\varepsilon_0}E=3ε0​ρr​

So Statement 2 is true.

  1. Potential difference between centre and surface

We use Vc−Vs=∫0RE(r) drV_c-V_s=\int_0^R E(r)\,drVc​−Vs​=∫0R​E(r)dr

because potential decreases outward.

Substitute E(r)=ρr3ε0E(r)=\dfrac{\rho r}{3\varepsilon_0}E(r)=3ε0​ρr​:

=\frac{\rho}{3\varepsilon_0}\int_0^R r\,dr =\frac{\rho}{3\varepsilon_0}\cdot \frac{R^2}{2}$$ Thus, $$V_c-V_s=\frac{\rho R^2}{6\varepsilon_0}$$ If a charge $q$ is moved from centre to surface, change in potential energy is $$\Delta U = q(V_s-V_c)=-q(V_c-V_s)=-\frac{q\rho R^2}{6\varepsilon_0}$$ Its magnitude is $$\left|\Delta U\right|=\frac{q\rho R^2}{6\varepsilon_0}$$ This is **not** equal to $\dfrac{q\rho}{3\varepsilon_0}$. Also, the given expression is dimensionally incorrect because it has no factor of $R^2$. Therefore **Statement 1 is false**. --- 4. **Evaluate the options** - **A:** Statement 1 true, Statement 2 true, but no explanation — false. - **B:** Statement 1 true, Statement 2 false — false. - **C:** Statement 1 false, Statement 2 true — **correct**. - **D:** Statement 1 true, Statement 2 true, and explanation — false. --- 5. **Final answer** The correct option is: $$\boxed{\text{C}}$$
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