Two identical particle s of mass m and charge q are shot at each other from a very great distance with an initial speed upsilon The distance of closest approach of the se charges is

Question

Two identical particle s of mass m and charge q are shot at each other from a very great distance with an initial speed upsilon The distance of closest approach of the se charges is (A) (q2/8πεo m upsilon2) (B) (q2/4πεo m upsilon2) (C) (q2/2πεo m upsilon2) (D) 0

Tardigrade Admin · Accepted Answer

At distance of closest approach, Total initial

K E =

Total final

PE

∴ \frac{1}{2} m υ^{2} + \frac{1}{2} m υ^{2} = \frac{k q ^{2}}{r}

\Rightarrow m υ^{2} = \frac{q ^{2}}{4 π ε _{o} r}

\Rightarrow r = \frac{q ^{2}}{4 π ε _{0} m v ^{2}}

Q. Two identical particle $s$ of mass $m$ and charge $q$ are shot at each other from a very great distance with an initial speed $υ$ The distance of closest approach of the se charges is

$\frac{q ^{2}}{8 π ε _{o} m υ ^{2}}$

$\frac{q ^{2}}{4 π ε _{o} m υ ^{2}}$

$\frac{q ^{2}}{2 π ε _{o} m υ ^{2}}$

$0$

At distance of closest approach, Total initial $K E =$ Total final $PE$
$∴ \frac{1}{2} m υ^{2} + \frac{1}{2} m υ^{2} = \frac{k q ^{2}}{r}$

$\Rightarrow m υ^{2} = \frac{q ^{2}}{4 π ε _{o} r}$
$\Rightarrow r = \frac{q ^{2}}{4 π ε _{0} m v ^{2}}$

Q. Two identical particle s of mass m and charge q are shot at each other from a very great distance with an initial speed υ The distance of closest approach of the se charges is

8πεo​mυ2q2​

4πεo​mυ2q2​

2πεo​mυ2q2​

0

At distance of closest approach, Total initial KE= Total final PE ∴21​mυ2+21​mυ2=rkq2​ ⇒mυ2=4πεo​rq2​ ⇒r=4πε0​mv2q2​

Q. Two identical particle $s$ of mass $m$ and charge $q$ are shot at each other from a very great distance with an initial speed $υ$ The distance of closest approach of the se charges is

$\frac{q ^{2}}{8 π ε _{o} m υ ^{2}}$

$\frac{q ^{2}}{4 π ε _{o} m υ ^{2}}$

$\frac{q ^{2}}{2 π ε _{o} m υ ^{2}}$

$0$

At distance of closest approach, Total initial $K E =$ Total final $PE$
$∴ \frac{1}{2} m υ^{2} + \frac{1}{2} m υ^{2} = \frac{k q ^{2}}{r}$

$\Rightarrow m υ^{2} = \frac{q ^{2}}{4 π ε _{o} r}$
$\Rightarrow r = \frac{q ^{2}}{4 π ε _{0} m v ^{2}}$