Question

A charge q = 10-6 C of mass 2 g is free to move when released at a distance 'a' from the fixed charge Q. Calculate its speed, when it reached to a distance b:- [Assume a = 1 m, b = 10 m, Q = 10-3C]

• A. 90 m/s
• B. 9 m/s
• C. 3 m/s
• D. 30 m/s

Answer 1

Answer: (A)

90 m/s

Answer 2

The problem can be solved using the principle of conservation of mechanical energy. The system consists of a fixed charge $Q$ and a movable charge $q$ of mass $m$. The electrostatic force is conservative, so the total mechanical energy (kinetic energy + potential energy) of the movable charge is conserved.

Initial state: The charge $q$ is at a distance $a$ from the fixed charge $Q$ and is released from rest.

Initial kinetic energy, $K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} m (0)^2 = 0$.

Initial potential energy, $U_i = \frac{k Q q}{a}$, where $k = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant.

Final state: The charge $q$ reaches a distance $b$ from the fixed charge $Q$. Let its speed at this distance be $v_f$.

Final kinetic energy, $K_f = \frac{1}{2} m v_f^2$.

Final potential energy, $U_f = \frac{k Q q}{b}$.

By conservation of mechanical energy:

$K_i + U_i = K_f + U_f$

$0 + \frac{k Q q}{a} = \frac{1}{2} m v_f^2 + \frac{k Q q}{b}$

Rearranging the equation to solve for $v_f$:

$\frac{1}{2} m v_f^2 = \frac{k Q q}{a} - \frac{k Q q}{b} = k Q q \left( \frac{1}{a} - \frac{1}{b} \right)$

$v_f^2 = \frac{2 k Q q}{m} \left( \frac{1}{a} - \frac{1}{b} \right)$

$v_f = \sqrt{\frac{2 k Q q}{m} \left( \frac{1}{a} - \frac{1}{b} \right)}$

Given values are:

$q = 10^{-6}$ C

$m = 2$ g $= 2 \times 10^{-3}$ kg

$a = 1$ m

$b = 10$ m

$Q = 10^{-3}$ C

$k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$

Substitute these values into the formula for $v_f$:

$v_f = \sqrt{\frac{2 \times (9 \times 10^9) \times (10^{-3}) \times (10^{-6})}{2 \times 10^{-3}} \left( \frac{1}{1} - \frac{1}{10} \right)}$

$v_f = \sqrt{\frac{18 \times 10^{9-3-6}}{2 \times 10^{-3}} \left( 1 - 0.1 \right)}$

$v_f = \sqrt{\frac{18 \times 10^0}{2 \times 10^{-3}} \times 0.9}$

$v_f = \sqrt{\frac{9}{10^{-3}} \times 0.9}$

$v_f = \sqrt{9 \times 10^3 \times 0.9}$

$v_f = \sqrt{9000 \times 0.9}$

$v_f = \sqrt{8100}$

$v_f = 90$ m/s

The speed of the charge when it reached to a distance $b$ is 90 m/s.