Question

Consider an infinite wire having linear charge density 8µC/m. A point charge 1nC having mass 2.8 mg is released from rest at a perpendicular distance 3 m from the wire. Find the velocity (in m/s) of the point charge when it is at a distance 12 m from the wire. (Take In 2 = 0.7).

Answer 1

12

Answer 2

The electric field due to an infinite line charge with linear charge density $\lambda$ at a perpendicular distance $r$ from the wire is given by $E = \frac{\lambda}{2\pi\epsilon_0 r}$. The direction of the field is radially outwards if $\lambda$ is positive.

The force on a point charge $q$ in this electric field is $F = qE = q \frac{\lambda}{2\pi\epsilon_0 r}$. Since both $\lambda$ and $q$ are positive, the force is directed radially outwards, away from the wire. This force is conservative, so we can use the principle of conservation of mechanical energy.

The potential energy of the point charge $q$ at a distance $r$ from the wire is $PE(r) = qV(r)$, where $V(r)$ is the electric potential at distance $r$. The potential difference between two points at distances $r_1$ and $r_2$ from the wire is given by:

$V(r_2) - V(r_1) = -\int_{r_1}^{r_2} \vec{E} \cdot d\vec{l}$

Since the motion is radial, $\vec{E} \cdot d\vec{l} = E dr$.

$V(r_2) - V(r_1) = -\int_{r_1}^{r_2} \frac{\lambda}{2\pi\epsilon_0 r} dr = -\frac{\lambda}{2\pi\epsilon_0} [\ln r]_{r_1}^{r_2} = -\frac{\lambda}{2\pi\epsilon_0} (\ln r_2 - \ln r_1) = \frac{\lambda}{2\pi\epsilon_0} (\ln r_1 - \ln r_2) = \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{r_1}{r_2}\right)$.

The change in potential energy as the charge moves from $r_i$ to $r_f$ is:

$\Delta PE = PE_f - PE_i = q(V(r_f) - V(r_i)) = q \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{r_i}{r_f}\right)$.

According to the conservation of mechanical energy, the change in kinetic energy plus the change in potential energy is zero:

$\Delta KE + \Delta PE = 0$

$KE_f - KE_i = - \Delta PE = -q \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{r_i}{r_f}\right) = q \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{r_f}{r_i}\right)$.

The point charge is released from rest at $r_i = 3$ m, so its initial velocity $v_i = 0$ and initial kinetic energy $KE_i = \frac{1}{2} m v_i^2 = 0$. The final distance is $r_f = 12$ m, and let the final velocity be $v_f$. The final kinetic energy is $KE_f = \frac{1}{2} m v_f^2$.

Substituting these into the energy conservation equation:

$\frac{1}{2} m v_f^2 - 0 = q \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{r_f}{r_i}\right)$.

$\frac{1}{2} m v_f^2 = q \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{12}{3}\right) = q \frac{\lambda}{2\pi\epsilon_0} \ln(4)$.

We are given: $\lambda = 8 \, \mu C/m = 8 \times 10^{-6} \, C/m$. $q = 1 \, nC = 1 \times 10^{-9} \, C$. $m = 2.8 \, mg = 2.8 \times 10^{-6} \, kg$. $\ln 2 = 0.7$. Note that $\ln 4 = \ln(2^2) = 2 \ln 2 = 2 \times 0.7 = 1.4$. The constant $\frac{1}{2\pi\epsilon_0}$ can be written in terms of Coulomb's constant $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, Nm^2/C^2$. So, $\frac{1}{2\pi\epsilon_0} = 2 \times \frac{1}{4\pi\epsilon_0} = 2k = 2 \times 9 \times 10^9 = 18 \times 10^9 \, Nm^2/C^2$.

Substitute the values into the equation for $v_f^2$:

$\frac{1}{2} (2.8 \times 10^{-6}) v_f^2 = (1 \times 10^{-9}) (8 \times 10^{-6}) (18 \times 10^9) (1.4)$.

$(1.4 \times 10^{-6}) v_f^2 = (8 \times 18 \times 1.4) \times (10^{-9} \times 10^{-6} \times 10^9)$.

$(1.4 \times 10^{-6}) v_f^2 = (8 \times 18 \times 1.4) \times 10^{-6}$.

Divide both sides by $(1.4 \times 10^{-6})$:

$v_f^2 = 8 \times 18$.

$v_f^2 = 144$.

$v_f = \sqrt{144} = 12$.

The velocity of the point charge when it is at a distance 12 m from the wire is 12 m/s.