2. Electric Potential and Capacitance
medium

Two electrons are moving towards each other, each with a velocity of $10^6 \,m / s$. What will be closest distance of approach between them is ......... $m$

A

$1.53 \times 10^{-8}$

B

$2.53 \times 10^{-10}$

C

$2.53 \times 10^{-6}$

D

$0$

Solution

(b)

$2 . \frac{1}{2}\left(9.1 \times 10^{-31}\right)\left(10^6\right)^2=\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{r}$

$9.1 \times 10^{-19}=\frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{r}$

$r=2.56 \times 10^{-10} \,m$

Standard 12
Physics

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