A particle ' $A$ ' has charge ' $+q$ ' and a particle ' $B$ '. has charge ' $+4 q$ '. Each has same mass ' m…
A particle ' $A$ ' has charge ' $+q$ ' and a particle ' $B$ '. has charge ' $+4 q$ '. Each has same mass ' m '. When they are allowed to fall from rest through the same potential, the ratio of their speeds will become (particle A to particle B)
$2: 1$
$1: 2$
$1: 4$
$4: 1$
Solution
Electric force on charged particle placed in a electric field.
$\begin{aligned}
& F=q E \\
& F_A=+q E \\
& F_A=m_A a_A \\
\therefore \quad a_A & =\frac{q E}{m} \\
F_B & =+4 q E \\
F_B & =m_B a_B \\
\quad & a_B=\frac{4 q E}{m} \\
\therefore \quad & a_A=\frac{a_B}{4}...(i)
\end{aligned}$ As both the bodies are falling from rest and cover same distance.
$\begin{aligned}
& V_A^2=0+2 a_A x \\
& V_B^2=0+2 a_B x
\end{aligned}$
$\begin{array}{ll}
\therefore \quad\left(\frac{V_A}{V_B}\right)^2=\frac{a_A}{a_B} \\
& \quad\left(\frac{V_A}{V_B}\right)^2=\frac{a_B}{4 \times a_B} \\
\therefore \quad & \frac{V_A}{V_B}=\frac{1}{2}
\end{array}$
...[From (i)]