Electrostatics> two oppositely charged rings of radius r ...

two oppositely charged rings of radius r are separated by a distance 6r. an electron is placed at the centre of the positively charged ring. what min velocity should be given to the electron so that it reaches the centre of the negatively charged ring

drishti , 7 Years ago

Grade 12

1 Answers

Eshan

Last Activity: 7 Years ago

Potential energy of electron when kept at the center of postively charged ring= $\dfrac{kQ(-e)}{r}+\dfrac{k(-Q)(-e)}{\sqrt{(6r)^2+(r)^2}}$
Potential energy of electron when it reaches at the center of negatively charged ring=
$\dfrac{kQ(-e)}{\sqrt{(6r)^2+(r)^2}}+\dfrac{k(-Q)(-e)}{r}$

The gain in potential energy must come at expense of loss of kinetic energy
$\implies \dfrac{kQ(-e)}{\sqrt{(6r)^2+(r)^2}}+\dfrac{k(-Q)(-e)}{r}-(\dfrac{kQ(-e)}{r}+\dfrac{k(-Q)(-e)}{\sqrt{(6r)^2+(r)^2}})=\dfrac{1}{2}mv^2$
$\implies kQe(-\dfrac{1}{\sqrt{37}r}+\dfrac{1}{r}+\dfrac{1}{r}-\dfrac{1}{\sqrt{37}r})=\dfrac{1}{2}mv^2$
$\implies v=\sqrt{\dfrac{4kQe}{m}(\dfrac{1}{r}-\dfrac{1}{\sqrt{37}r})}$