There is a uniform spherically symmetric surface charge density at a distance R0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V (R(t)) of the distribution as a function of its instantaneous radius R (t) is :

Electrostatics PYQ — JEE Main 2019 · 12 Jan · Shift 1 · Q73 with Solution

View written solutionFree

Correct answer: B

Model the system

A thin spherical shell of total charge $Q$ and radius $R(t)$ has uniform surface charge density at every instant.

Because of spherical symmetry:

the electric field inside the shell is zero,
the field just outside is as if all charge were concentrated at the center.

As the shell expands due to mutual repulsion, every small element of the shell moves radially outward.

Electrostatic potential energy of a spherical shell

For a spherical shell of radius $R$ , the electrostatic self-energy is

=\frac{1}{8\pi\varepsilon_0}\frac{Q^2}{R}.$$ So, $$U(R)=k\frac{Q^2}{2R}, \qquad \text{where } k=\frac{1}{4\pi\varepsilon_0}.$$ Initially, at radius $R_0$, the shell is at rest, so kinetic energy is zero. --- 3. **Apply conservation of energy** Let total mass of the shell be $M$ and instantaneous speed of the shell be $V(R)$. Initially: $$E_i=U(R_0)=k\frac{Q^2}{2R_0}.$$ At radius $R$: $$E=\frac{1}{2}MV^2 + k\frac{Q^2}{2R}.$$ Using conservation of energy, $$k\frac{Q^2}{2R_0}=\frac{1}{2}MV^2+k\frac{Q^2}{2R}.$$ Therefore, $$\frac{1}{2}MV^2 = k\frac{Q^2}{2}\left(\frac{1}{R_0}-\frac{1}{R}\right).$$ Hence, $$V^2 = \frac{kQ^2}{M}\left(\frac{1}{R_0}-\frac{1}{R}\right),$$ so $$V(R)=\sqrt{\frac{kQ^2}{M}\left(\frac{1}{R_0}-\frac{1}{R}\right)}.$$ --- 4. **Study the shape of $V$ vs $R$** From the formula, $$V(R)=A\sqrt{\frac{1}{R_0}-\frac{1}{R}},$$ where $A=\sqrt{\frac{kQ^2}{M}}$ is constant. Now observe: - At $R=R_0$, $$V=0.$$ - As $R$ increases, $1/R$ decreases, so $V$ increases. - As $R\to\infty$, $$V\to A\sqrt{\frac{1}{R_0}},$$ i.e. the speed approaches a **finite limiting value**. So the graph: - starts from $(R_0,0)$, - rises with increasing $R$, - gradually flattens, - approaches a horizontal asymptote. Also, $$\frac{dV}{dR}=\frac{A}{2V}\cdot \frac{1}{R^2}>0,$$ so it is increasing, and the slope decreases as $R$ grows large, consistent with saturation. --- 5. **Conclusion** The correct graph is the one that increases from zero at $R=R_0$ and approaches a constant value asymptotically. That corresponds to **Option B**. --- 6. **Comparison with stored answer** Stored correct answer: **B** Derived answer: **B** So they agree.

More from Electrostatics

View written solutionFree

Correct answer: B

Model the system

A thin spherical shell of total charge $Q$ and radius $R(t)$ has uniform surface charge density at every instant.

Because of spherical symmetry:

the electric field inside the shell is zero,
the field just outside is as if all charge were concentrated at the center.

As the shell expands due to mutual repulsion, every small element of the shell moves radially outward.

Electrostatic potential energy of a spherical shell

For a spherical shell of radius $R$ , the electrostatic self-energy is

=\frac{1}{8\pi\varepsilon_0}\frac{Q^2}{R}.$$ So, $$U(R)=k\frac{Q^2}{2R}, \qquad \text{where } k=\frac{1}{4\pi\varepsilon_0}.$$ Initially, at radius $R_0$, the shell is at rest, so kinetic energy is zero. --- 3. **Apply conservation of energy** Let total mass of the shell be $M$ and instantaneous speed of the shell be $V(R)$. Initially: $$E_i=U(R_0)=k\frac{Q^2}{2R_0}.$$ At radius $R$: $$E=\frac{1}{2}MV^2 + k\frac{Q^2}{2R}.$$ Using conservation of energy, $$k\frac{Q^2}{2R_0}=\frac{1}{2}MV^2+k\frac{Q^2}{2R}.$$ Therefore, $$\frac{1}{2}MV^2 = k\frac{Q^2}{2}\left(\frac{1}{R_0}-\frac{1}{R}\right).$$ Hence, $$V^2 = \frac{kQ^2}{M}\left(\frac{1}{R_0}-\frac{1}{R}\right),$$ so $$V(R)=\sqrt{\frac{kQ^2}{M}\left(\frac{1}{R_0}-\frac{1}{R}\right)}.$$ --- 4. **Study the shape of $V$ vs $R$** From the formula, $$V(R)=A\sqrt{\frac{1}{R_0}-\frac{1}{R}},$$ where $A=\sqrt{\frac{kQ^2}{M}}$ is constant. Now observe: - At $R=R_0$, $$V=0.$$ - As $R$ increases, $1/R$ decreases, so $V$ increases. - As $R\to\infty$, $$V\to A\sqrt{\frac{1}{R_0}},$$ i.e. the speed approaches a **finite limiting value**. So the graph: - starts from $(R_0,0)$, - rises with increasing $R$, - gradually flattens, - approaches a horizontal asymptote. Also, $$\frac{dV}{dR}=\frac{A}{2V}\cdot \frac{1}{R^2}>0,$$ so it is increasing, and the slope decreases as $R$ grows large, consistent with saturation. --- 5. **Conclusion** The correct graph is the one that increases from zero at $R=R_0$ and approaches a constant value asymptotically. That corresponds to **Option B**. --- 6. **Comparison with stored answer** Stored correct answer: **B** Derived answer: **B** So they agree.

Electrostatics question

Electrostatics question

More from Electrostatics

Electrostatics question

Electrostatics question

More from Electrostatics