Question

A ball of mass $1g$ and charge ${10}^{-8}C$ moves from $A$, where potential is $600V$, to $B$, where potential is $0$. The velocity of the ball at B is $20cm/s$. The velocity of the ball at point $A$ must have been

• A. $16.8cm/s$
• B. $22.8cm/s$
• C. $228cm/s$
• D. $168cm/s$

Answer 1

The correct option is B $22.8cm/s$

Given a ball of $m=1g={10}^{-3}kg$

charge on ball, $q={10}^{-8}C$

potential at $A;v_A=600v$

potential at $B;v_B=0$

velocity of ball at $ B;{ v }_{ B }=20cm/s

we have to find velocity of ball at $A$

$ex.v_A=?$

so; we use conservation of energy,

change in kinetic energy= change in potential energy.

$\frac{1}{2}m({v_A}^2-{v_B}^2)=Q(v_A-v_B)\frac{1}{2}\times {10}^{-3}({v_A}^2-{\left(0.2\right)}^2)={10}^{-8}\times (600-0){v_A}^2-{\left(0.2\right)}^2=6\times {10}^{-6}+2\times {10}^3=12\times {10}^{-3}{v_A}^2=12\times {10}^{-3}+4\times {10}^3{v}_A=22.8{cms}^{-1}$

hence the answer is option (b)