Question

# Two identical coaxial rings each of radius R are separated by a distance of $$\sqrt 3 R$$, They are uniformly charged with charges +Q and -Q respectively. The minimum kinetic energy with which a charged particle (charge + q) should be projected from the center of the negatively charged ring along the axis of the rings such that it reaches the center of the positively charged ring is :

A
Qq4πε0R
B
Qq2πε0R
C
Qq8πε0R
D
3Qq4πε0R

Solution

## The correct option is A $$\displaystyle \frac{Qq}{4 \pi \varepsilon_0 R}$$Use energy conservation,                                        $$U_{i}\; =\; U_{f}$$                                        $$\dfrac{1}{2}mv^{2}+\dfrac{KQq}{\sqrt{(\sqrt{3R}^{2}+R^{2})}}+\dfrac{K(-Q)q}{\sqrt{\sqrt{0}^{2}+R^{2}}}\; =\; \dfrac{KQq}{\sqrt{(\sqrt{0}^{2}+R^{2})}}+\dfrac{K(-Q)q}{\sqrt{(\sqrt{3R}^{2}+R^{2})}}$$                                       $$\Rightarrow \dfrac{1}{2}mv^{2}\; =\; \dfrac{KQq}{R}$$Physics

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