Question

Two point charges each $50\mu C$ are fixed on the y-axis. Another charged particle having charge Q1 and a mass 20 gm moving with a speed of $20m{s}^{-1}$. Find the speed of charged particle when it reaches the origin? Also, find the distance of particle when Kinetic energy is 0.

Answer 1

Step 1: Given

Charge of two point charges= 50μC

Mass of another charge, $Q_1=20gm$

Speed of $Q_1=20m{s}^{-1}$

Step 2: Formula used and calculation

Let the speed of the particle at origin be v. Applying the energy conservation at A and O.

Initial energy=final energy

$K_A+U_A=K_o+U_o$

$\frac{1}{2}m{v}_A^2+(-Q)V_A=\frac{1}{2}m{v}_o^2+(-Q)V_o$

$\frac{1}{2}m{v}_A^2-\frac{2K{Q}^2}{5}=\frac{1}{2}m{v}_o^2-\frac{2K{Q}^2}{4}$

$\frac{1}{2}m{v}_o^2=\frac{1}{2}\times 20\times {10}^{-3}\times 20\times 20+9\times {10}^9\times {(50\times {10}^{-6})}^2(\frac{1}{4}-\frac{1}{5})$

$\frac{1}{2}m{v}_o^2=4+9\times 2.5\times \frac{1}{20}$

$v_o^2=225$

$v_o=25m{s}^{-1}$

Let at point B kinetic energy is 0.

Applying the energy conservation at A and B.

$K_A+U_A=K_B+U_B$

$\frac{1}{2}m{v}_A^2+\left(\frac{-2K{Q}^2}{5}\right)=0+(-\frac{K{Q}^2}{\sqrt{4^2+x^2}})$

$\frac{1}{2}m{v}_A^2=\frac{2K{Q}^2}{5}-\frac{K{Q}^2}{\sqrt{4^2+x^2}}$

$\frac{1}{2}m{v}_A^2=9\times {10}^9\times {(50\times {10}^{-6})}^2(\frac{2}{5}-\frac{1}{\sqrt{4^2+x^2}})$

$4=22.5\times \frac{2}{5}-\frac{22.5}{\sqrt{4^2+x^2}}$

$\frac{22.5}{\sqrt{4^2+x^2}}=9-4$

$\sqrt{4^2+x^2}=4.5$

$x^2={4.5}^2-4^2$

$x=\sqrt{4.25}$

$x=2.06m$