Q89 / 90 · JEE Main 2012 (Offline) · Physics

This question has statement $1$ and statement $2$ . Of the four choices given after the statements, choose the one that best describes the two statements. An insulating solid sphere of radius $\mathrm{R}$ has a uniformly positive charge density $\rho$. As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinity is zero. Statement $1$: When a charge $q$ is taken from the centre to the surface of the sphere, its potential energy changes by $\frac{\mathrm{qp}}{3 \varepsilon_0}$ Statement $2$: The electric field at a distance $r(r < R)$ from the centre of the sphere is $\frac{\rho r}{3 \varepsilon_0}$

  • A
    Statement $1$ is true, Statement $2$ is true, Statement $2$ is not the correct explanation for statement $1$.
  • B
    Statement $1$ is true, Statement $2$ is false
  • Correct answer
    C
    Statement $1$ is false, Statement $2$ is true
  • D
    Statement $1$ is true, Statement $2$ is the correct explanation for statement $1$

Solution

$\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d}} \mathrm{A}=\frac{1}{\varepsilon_0}\left(\rho \times \frac{4}{3} \pi \mathrm{r}^3\right)$ $\mathrm{E}=\frac{\rho \mathrm{r}}{3 \varepsilon_0}$ Statement $2$ is correct $\Delta \mathrm{PE}=\left(\mathrm{V}_{\text {sur }}-\mathrm{V}_{\text {cent }}\right) \mathrm{q}=-\frac{\mathrm{q}}{6 \varepsilon_0} \rho \mathrm{R}^2$ Statement $1$ is incorrect

Hence, the correct answer is option: 'C'.

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