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Question

If we throw the charged particle just right of the $v$ center of the tunnel, the particle will cross the tunnel. Hence applying conservation of $M E$

Vedclass · Accepted Answer

$[\rho R^2/4m\varepsilon_0]^{1/2}$

Vedclass · Answer

If we throw the charged particle just right of the $v$ center of the tunnel, the particle will cross the tunnel. Hence applying conservation of $M E$ between start point and center of tunnel, $\Delta K+\Delta U=0$

or $\left(0+\frac{1}{2} m v^{2}\right)+q\left(V_{f}-V_{i}\right)=0$

or $V_{f}=\frac{V_{s}}{2}\left(3-\frac{r^{2}}{R^{2}}\right)=\frac{p R^{2}}{6 \varepsilon_{0}}\left(3-\frac{r^{2}}{R^{2}}\right)$

Hence $r=\frac{R}{2}$

$V_{f}=\frac{p R^{2}}{6 \varepsilon_{0}} \cdot\left(3-\frac{R^{2}}{4 R^{2}}\right)=\frac{11 p R^{2}}{24 \varepsilon_{0}}$

$V_{i}=\left(\frac{p R^{2}}{3 \varepsilon_{90}}\right)$

$\frac{1}{2} m v^{2}=1\left[\frac{11 p R^{2}}{24 \varepsilon_{0}}-\frac{p R^{2}}{3 \varepsilon_{90}}\right]=\frac{p R^{2}}{3 \varepsilon_{0}}\left[\frac{11}{24}-\frac{1}{3}\right]$

$=\frac{p R^{2}}{8 \varepsilon_{0}}$

or $V=\left(\frac{p R^{2}}{4 m \varepsilon_{0}}\right)^{1 / 2}$

Hence, velocity should be slightly greater than $V$.

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