Question

A unit positive point charge of mass m is projected with a velocity v inside the tunnel as shown. The tunnel has been made inside a uniformly charged non conducting sphere of volume charge density ρ. The minimum velocity with which the point charge should be projected such that it can reach the opposite end of the tunnel, is equal to $\sqrt{\frac{\rho R}{m\epsilon_0}}x$. Find x.

Answer 1

1/sqrt{3}

Answer 2

The problem describes a unit positive point charge of mass $m$ projected with velocity $v$ inside a tunnel made in a uniformly charged non-conducting sphere of radius $R$ and volume charge density $\rho$. The tunnel is along a diameter. The charge is projected from one end of the tunnel (say, $x=R$) towards the center, and we need to find the minimum velocity $v$ such that it reaches the opposite end of the tunnel (i.e., $x=-R$).

1. Electric Field and Force: For a uniformly charged non-conducting sphere, the electric field inside at a distance $x$ from the center is given by: $E = \frac{\rho x}{3\epsilon_0}$ The direction of the electric field is radially outwards if $\rho$ is positive. The force on a unit positive charge ($q=1$) at position $x$ is: $F = qE = \frac{\rho x}{3\epsilon_0}$ This force is always directed away from the center (repulsive).

2. Equation of Motion: Let the center of the sphere be the origin ($x=0$). The charge is projected from $x=R$ towards the center. So, the initial velocity is $v_x = -v$. The equation of motion for the charge is: $m \frac{d^2x}{dt^2} = -F_x$ (since the force is directed away from the center, it opposes the inward motion) $m \frac{d^2x}{dt^2} = -\frac{\rho x}{3\epsilon_0}$ This is incorrect. The force is $F_x = \frac{\rho x}{3\epsilon_0}$. So the equation of motion is $m \frac{d^2x}{dt^2} = F_x = \frac{\rho x}{3\epsilon_0}$. $m \frac{d^2x}{dt^2} - \frac{\rho x}{3\epsilon_0} = 0$ This is a second-order linear differential equation of the form $\frac{d^2x}{dt^2} - \omega^2 x = 0$, where $\omega^2 = \frac{\rho}{3m\epsilon_0}$. The general solution is $x(t) = C_1 e^{\omega t} + C_2 e^{-\omega t}$.

3. Applying Initial Conditions: At $t=0$, $x(0) = R$. So, $R = C_1 + C_2$. The velocity is $v_x(t) = \frac{dx}{dt} = \omega C_1 e^{\omega t} - \omega C_2 e^{-\omega t}$. At $t=0$, $v_x(0) = -v$ (since the charge is projected towards the center). So, $-v = \omega C_1 - \omega C_2 \implies C_1 - C_2 = -v/\omega$.

   Solving for $C_1$ and $C_2$: Adding the two equations: $2C_1 = R - v/\omega \implies C_1 = \frac{1}{2} (R - v/\omega)$. Subtracting the second from the first: $2C_2 = R + v/\omega \implies C_2 = \frac{1}{2} (R + v/\omega)$.

   Thus, the position as a function of time is: $x(t) = \frac{1}{2} (R - v/\omega) e^{\omega t} + \frac{1}{2} (R + v/\omega) e^{-\omega t}$.

4. Condition for Reaching the Opposite End: The force on the charge is always repulsive from the center. If the charge is at $x>0$, the force is in the $+x$ direction. If the charge is at $x<0$, the force is in the $-x$ direction. The charge is projected from $x=R$ towards $x=0$. As it moves towards $x=0$, the repulsive force opposes its motion. For the charge to reach the opposite end ($x=-R$), it must at least reach $x=0$. Once it crosses $x=0$ (i.e., enters the region $x<0$), the force will be in the $-x$ direction, accelerating it towards $x=-R$. Therefore, the minimum velocity required is such that the charge just reaches $x=0$. This means its velocity becomes zero at $x=0$, or it just passes through $x=0$. If $v_x(t)$ becomes zero at some $x > 0$, the particle will turn back and not reach $x=0$. The velocity becomes zero when $v_x(t) = 0$: $\omega C_1 e^{\omega t} - \omega C_2 e^{-\omega t} = 0$ $C_1 e^{\omega t} = C_2 e^{-\omega t}$ $e^{2\omega t} = C_2/C_1$ For a real time $t$, $C_2/C_1$ must be positive. Since $C_2 = \frac{1}{2} (R + v/\omega)$ is always positive (as $R, v, \omega$ are positive), $C_1$ must also be positive. $C_1 > 0 \implies \frac{1}{2} (R - v/\omega) > 0 \implies R - v/\omega > 0 \implies R > v/\omega \implies v < R\omega$. If $v < R\omega$, the particle will stop at a positive $x$ value and turn back. The stopping position is $x_{stop} = 2\sqrt{C_1 C_2} = 2\sqrt{\frac{1}{4}(R - v/\omega)(R + v/\omega)} = \sqrt{R^2 - (v/\omega)^2}$. For the particle to reach $x=0$ or beyond, it must not stop at $x>0$. This means $x_{stop}$ must be less than or equal to $0$. $\sqrt{R^2 - (v/\omega)^2} \le 0$ This implies $R^2 - (v/\omega)^2 \le 0$ $R^2 \le (v/\omega)^2$ $v^2 \ge R^2 \omega^2$ $v \ge R\omega$

   The minimum velocity is when $v = R\omega$. Substituting $\omega = \sqrt{\frac{\rho}{3m\epsilon_0}}$: $v_{min} = R \sqrt{\frac{\rho}{3m\epsilon_0}} = \sqrt{\frac{\rho R^2}{3m\epsilon_0}} = \sqrt{\frac{\rho R}{m\epsilon_0}} \sqrt{\frac{R}{3}}$. This doesn't match the form given in the question.

   Let's recheck the force direction. If the charge is positive, and $\rho$ is positive, the electric field $\vec{E} = \frac{\rho \vec{r}}{3\epsilon_0}$ is radially outwards. The force $\vec{F} = q\vec{E} = \frac{\rho \vec{r}}{3\epsilon_0}$ is also radially outwards. If the tunnel is along the x-axis, and the charge is at $x$, then $\vec{r} = x\hat{i}$. So, $\vec{F} = \frac{\rho x}{3\epsilon_0}\hat{i}$. The equation of motion is $m \frac{d^2x}{dt^2} = \frac{\rho x}{3\epsilon_0}$. This is correct. This is an unstable equilibrium at $x=0$.

   Let's re-verify the conditions. If $v = R\omega$, then $C_1 = \frac{1}{2}(R - R) = 0$. In this case, $x(t) = \frac{1}{2} (R + R) e^{-\omega t} = R e^{-\omega t}$. And $v_x(t) = -\omega R e^{-\omega t}$. As $t \to \infty$, $x(t) \to 0$ and $v_x(t) \to 0$. This means the particle asymptotically approaches $x=0$. It never strictly reaches $x=0$ in finite time. However, in physics problems, "just reaches" often implies this boundary condition. If it reaches $x=0$, it will then be accelerated to $x=-R$. So, $v_{min} = R\omega$ is the correct condition.

   $v_{min} = R \sqrt{\frac{\rho}{3m\epsilon_0}} = \sqrt{\frac{\rho R^2}{3m\epsilon_0}}$ The given form is $v = \sqrt{\frac{\rho R}{m\epsilon_0}}x$. Let's write $v_{min}$ as $\sqrt{\frac{\rho R}{m\epsilon_0}} \cdot \sqrt{\frac{R}{3}}$. Comparing this with $\sqrt{\frac{\rho R}{m\epsilon_0}}x$, we get $x = \sqrt{\frac{R}{3}}$.

   Wait, the 'x' in the question is a numerical value, not a variable 'R'. The question asks to "Find x". Let's look at the given expression: $\sqrt{\frac{\rho R}{m\epsilon_0}}x$. My derived expression is $v_{min} = \sqrt{\frac{\rho R^2}{3m\epsilon_0}}$. Comparing $\sqrt{\frac{\rho R^2}{3m\epsilon_0}}$ with $\sqrt{\frac{\rho R}{m\epsilon_0}}x$: $\sqrt{\frac{\rho R^2}{3m\epsilon_0}} = \sqrt{\frac{\rho R}{m\epsilon_0}} \cdot x$ Squaring both sides: $\frac{\rho R^2}{3m\epsilon_0} = \frac{\rho R}{m\epsilon_0} \cdot x^2$ $\frac{R}{3} = x^2$ $x = \sqrt{\frac{R}{3}}$.

   This implies $x$ is dependent on $R$. However, $x$ is usually a dimensionless constant in such problems. Let's check the units. $v_{min} = \sqrt{\frac{\rho R^2}{3m\epsilon_0}}$ Units: $\sqrt{\frac{(\text{C/m}^3) \cdot \text{m}^2}{\text{kg} \cdot (\text{C}^2/(\text{N}\cdot\text{m}^2))}} = \sqrt{\frac{\text{C/m}}{\text{kg} \cdot (\text{C}^2/(\text{N}\cdot\text{m}^2))}} = \sqrt{\frac{\text{N}\cdot\text{m}^2}{\text{kg}\cdot\text{m}\cdot\text{C}}} = \sqrt{\frac{\text{N}\cdot\text{m}}{\text{kg}\cdot\text{C}}} = \sqrt{\frac{\text{J}}{\text{kg}\cdot\text{C}}}$. This is not velocity. There must be a unit charge $q$ involved in force. Force $F = qE = q \frac{\rho x}{3\epsilon_0}$. $m \frac{d^2x}{dt^2} = q \frac{\rho x}{3\epsilon_0}$. $\omega^2 = \frac{q\rho}{3m\epsilon_0}$. The problem states "A unit positive point charge", so $q=1$ (in units of charge, if we are using SI, it's 1 Coulomb, but often it refers to a dimensionless factor). Let's assume $q=1$ C. Then $\omega^2 = \frac{\rho}{3m\epsilon_0}$ is dimensionally correct. Units of $\omega$: $\sqrt{\frac{\text{C/m}^3}{\text{kg} \cdot \text{C}^2/(\text{N}\cdot\text{m}^2)}} = \sqrt{\frac{\text{N}\cdot\text{m}^2}{\text{kg}\cdot\text{m}^3}} = \sqrt{\frac{\text{N}}{\text{kg}\cdot\text{m}}} = \sqrt{\text{s}^{-2}} = \text{s}^{-1}$. This is correct for angular frequency. So $v_{min} = R\omega = R \sqrt{\frac{\rho}{3m\epsilon_0}}$. Units: $\text{m} \cdot \text{s}^{-1} = \text{m/s}$. This is correct for velocity.

   Now, let's re-evaluate the comparison. $v_{min} = \sqrt{\frac{\rho R^2}{3m\epsilon_0}}$. The question states $v = \sqrt{\frac{\rho R}{m\epsilon_0}}x$. So, $\sqrt{\frac{\rho R^2}{3m\epsilon_0}} = \sqrt{\frac{\rho R}{m\epsilon_0}}x$. $\sqrt{\frac{\rho R}{m\epsilon_0}} \sqrt{\frac{R}{3}} = \sqrt{\frac{\rho R}{m\epsilon_0}}x$. Therefore, $x = \sqrt{\frac{R}{3}}$.

   This result means $x$ is not a pure number, but depends on $R$. This is highly unusual for a quantity asked to be "found" in such a format, suggesting $x$ should be a number. Could the question mean the radius of the sphere is 1 unit? No, it's R. Could the problem be interpreted differently? What if the tunnel is not along the diameter, but is a small tunnel at $R/2$? "The tunnel has been made inside a uniformly charged non conducting sphere of volume charge density ρ. The minimum velocity with which the point charge should be projected such that it can reach the opposite end of the tunnel, is equal to $\sqrt{\frac{\rho R}{m\epsilon_0}}x$." The diagram shows a tunnel along the diameter. A is on the surface. P is at R/2. The projection is from A. "Opposite end of the tunnel" means the other end of the diameter. So the interpretation of the motion from $x=R$ to $x=-R$ is correct.

   Let's check if there's any other interpretation of the electric field formula or the setup. The electric field inside a non-conducting sphere is $E = \frac{\rho r}{3\epsilon_0}$. This is standard. The force is $F = qE$. For unit positive charge, $q=1$. The equation of motion $m \frac{d^2x}{dt^2} = \frac{\rho x}{3\epsilon_0}$ is correct. The solution $x(t) = C_1 e^{\omega t} + C_2 e^{-\omega t}$ where $\omega = \sqrt{\frac{\rho}{3m\epsilon_0}}$ is correct. The initial conditions $x(0)=R, v_x(0)=-v$ are correct. The derivation of $C_1 = \frac{1}{2} (R - v/\omega)$ and $C_2 = \frac{1}{2} (R + v/\omega)$ is correct. The condition for reaching $x=0$ (or beyond) is $C_1 \le 0$. If $C_1 < 0$, then $R - v/\omega < 0 \implies v > R\omega$. In this case, $x(t)$ will always decrease and become negative, reaching $x=-R$. If $C_1 = 0$, then $v = R\omega$. In this case, $x(t) = R e^{-\omega t}$, which asymptotically approaches $0$. So the minimum velocity is $v_{min} = R\omega$.

   The result $x = \sqrt{R/3}$ seems robust given the standard physics. Perhaps the question is designed such that $R$ is implicitly set to some value like $R=3$? If $R=3$, then $x=\sqrt{3/3}=1$. Or maybe the question expects a numerical value for $x$, implying $R$ is a specific value in the problem setup, or $R$ should cancel out. Let's check if there is any mistake in reading the question or if the expression $\sqrt{\frac{\rho R}{m\epsilon_0}}x$ is a typo. If the question meant $\sqrt{\frac{\rho R^2}{m\epsilon_0}}x$, then $x = 1/\sqrt{3}$. If the question meant $\sqrt{\frac{\rho}{m\epsilon_0}}x$, then $x = \sqrt{R^2/3} = R/\sqrt{3}$.

   Given the options are usually numerical, $x=1/\sqrt{3}$ is the most plausible answer, assuming a typo in the question's given expression. The provided expression is $\sqrt{\frac{\rho R}{m\epsilon_0}}x$. My derived expression is $v_{min} = \sqrt{\frac{\rho R^2}{3m\epsilon_0}}$. If we are forced to find $x$ from the given form, then $x = \sqrt{\frac{R}{3}}$. However, if $x$ is expected to be a numerical value, then $R$ must be a specific numerical value (e.g., $R=3$ to make $x=1$). Without any information about $R$, and assuming $x$ must be a pure number, the only way is if there is a typo in the question's format.

   Let's proceed with the assumption that the question implies $x$ is a numerical constant, and the most common form for such problems would lead to $x=1/\sqrt{3}$. This implies the given expression should have been $\sqrt{\frac{\rho R^2}{m\epsilon_0}}x$.

   If we strictly follow the given question: $v_{min} = \sqrt{\frac{\rho R}{m\epsilon_0}}x$. We found $v_{min} = R\omega = R \sqrt{\frac{\rho}{3m\epsilon_0}}$. $R \sqrt{\frac{\rho}{3m\epsilon_0}} = \sqrt{\frac{\rho R}{m\epsilon_0}}x$. $\sqrt{R^2 \frac{\rho}{3m\epsilon_0}} = \sqrt{\frac{\rho R}{m\epsilon_0}}x$. $\sqrt{\frac{\rho R}{m\epsilon_0}} \sqrt{\frac{R}{3}} = \sqrt{\frac{\rho R}{m\epsilon_0}}x$. So $x = \sqrt{\frac{R}{3}}$.

   If $x$ is a numerical value, then $R$ must be a specific value. For example, if the question meant $R=3$ (in meters, assuming consistent units), then $x=1$. If the question is from a multiple choice exam, the options would usually clarify this.

   Let's assume the question expects a numerical constant for $x$. Consider the possibility that the radius of the sphere is $R_0$, and the point of projection is at distance $R$ from the center, where $R$ is some specific value related to $R_0$. But the diagram shows the sphere has radius $R$, and the projection is from the surface. So the starting point is at distance $R$ from the center.

   What if the problem meant the charge is projected from $x=R/2$ (point P in the diagram)? But the arrow for $v$ is at A, which is at the surface. "projected with a velocity v inside the tunnel as shown." The diagram shows v at A, on the surface.

   Let's assume the question implicitly implies that $R$ is such that $x$ becomes a specific numerical value. The most common numerical value in such problems is $1/\sqrt{3}$. This would happen if the expression in the question was $\sqrt{\frac{\rho R^2}{m\epsilon_0}}x$.

   However, if we are to strictly follow the given question: $v_{min} = \sqrt{\frac{\rho R}{m\epsilon_0}}x$. And our derived $v_{min} = \sqrt{\frac{\rho R^2}{3m\epsilon_0}}$. Then $x = \sqrt{\frac{R}{3}}$. This is the direct mathematical consequence. If $x$ is expected to be a numerical value, this problem is ill-posed without specifying $R$.

   However, in problems of this type, when they ask for 'x' in such a format, it usually implies that 'x' is a dimensionless constant. This means the units of $v_{min}$ must match the units of $\sqrt{\frac{\rho R}{m\epsilon_0}}$. Units of $\sqrt{\frac{\rho R}{m\epsilon_0}}$: $\sqrt{\frac{(\text{C/m}^3) \cdot \text{m}}{\text{kg} \cdot (\text{C}^2/(\text{N}\cdot\text{m}^2))}} = \sqrt{\frac{\text{C/m}^2}{\text{kg} \cdot \text{C}^2/(\text{N}\cdot\text{m}^2)}} = \sqrt{\frac{\text{N}\cdot\text{m}^2}{\text{kg}\cdot\text{m}^2}} = \sqrt{\frac{\text{N}}{\text{kg}}} = \sqrt{\text{m/s}^2} = \text{m}^{1/2}\text{s}^{-1}$. This is not a velocity (m/s). This means the factor $\sqrt{\frac{\rho R}{m\epsilon_0}}$ is NOT the correct unit for velocity. My derived $v_{min} = \sqrt{\frac{\rho R^2}{3m\epsilon_0}}$ has units of m/s. So, the given expression $\sqrt{\frac{\rho R}{m\epsilon_0}}x$ must be dimensionally consistent with velocity. For this to be true, $x$ must have units of $\text{m}^{1/2}$. This means $x$ is not a dimensionless constant.

   This clearly indicates a dimensional inconsistency in the problem statement's given form of $v$. Let's assume the question intended to give a dimensionally correct expression for $v_{min}$ of the form $v_{min} = \sqrt{\frac{\rho R^2}{m\epsilon_0}} \cdot (\text{dimensionless } x)$. In this case, $v_{min} = \sqrt{\frac{\rho R^2}{3m\epsilon_0}}$. So, $\sqrt{\frac{\rho R^2}{3m\epsilon_0}} = \sqrt{\frac{\rho R^2}{m\epsilon_0}}x$. $\frac{1}{\sqrt{3}} = x$. This is the most probable intended answer.

The final answer is $\boxed{1/\sqrt{3}}$.