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Question

(a)

Distances and velocities of charged particle given are as below.

Energy conservation between $A$ and $B$ gives,

$q V_A+\frac{1}{2} m u^2=

Vedclass · Accepted Answer

$\sqrt{3} u$

Vedclass · Answer

(a)

Distances and velocities of charged particle given are as below.

Energy conservation between $A$ and $B$ gives,

$q V_A+\frac{1}{2} m u^2=q V_B+\frac{1}{2} m(\sqrt{2} u)^2$

$\Rightarrow \quad q\left(V_A-V_B\right)=\frac{1}{2} m u^2$

$\Rightarrow \quad q \frac{\lambda}{2 \pi \varepsilon_0} \cdot \ln 2=\frac{1}{2} m u^2 \quad \dots(i)$

Now, energy conservation between $A$ and $C$ gives,

$q V_A +\frac{1}{2} m u^2=q V_C+\frac{1}{2} m v^2$

$\Rightarrow \frac{1}{2} m v^2 =q\left(V_A-V_C\right)+\frac{1}{2} m u^2$

$=q \frac{\lambda}{2 \pi \varepsilon_0} \cdot \ln 4+\frac{1}{2} m u^2$

$=\frac{2 q \lambda}{2 \pi \varepsilon_0} \cdot \ln 2+\frac{1}{2} m u^2$

Substituting value of $\left(\frac{q {\lambda}}{2 \pi \varepsilon_0} \ln 2\right)$ from Eq. $(i)$ in above equation, we have

$\frac{1}{2} m v^2=2\left(\frac{1}{2} m u^2\right)+\frac{1}{2} m u^2$

$\Rightarrow \quad v=\sqrt{3} u$

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