Question

In figure, a particle has mass $m=5g$ and charge $q^{\prime }=2\times {10}^{-9}C$ starts from rest at point $a$ and moves in a straight line to point $b$. What is its speed $v$ at point $b$?

• A. $2.65{cms}^{-1}$
• B. $3.65{cms}^{-1}$
• C. $4.65{cms}^{-1}$
• D. $5.65{cms}^{-1}$

Answer 1

The correct option is C $4.65{cms}^{-1}$

Potential at a point $p$ due to the charge $q$, $V_p=\frac{Kq}{r}$

Where, $K=\frac{1}{4\pi {ϵ}_o}=9\times {10}^9$

Total potential at point $a$, $V_a=\frac{9\times {10}^9\times 3\times {10}^{-9}}{{10}^{-2}}+\frac{9\times {10}^9\times (-3\times {10}^{-9})}{2\times {10}^{-2}}=13.5\times {10}^2$ V

Total potential at point $b$, $V_b=\frac{9\times {10}^9\times 3\times {10}^{-9}}{2\times {10}^{-2}}+\frac{9\times {10}^9\times (-3\times {10}^{-9})}{{10}^{-2}}=-13.5\times {10}^2$ V

Potential difference between point a and b, $V_{ab}=V_a-V_b=27\times {10}^2$ V

Work done in moving the charge $q^{\prime }$ from a to b, $W_{ab}=q^{\prime }{V}_{ab}$

$W_{ab}=2\times {10}^{-9}\times 2700=5.4\times {10}^{-6}$ J

From work-energy theorem : $W_{ab}=\Delta K.E$

$\Rightarrow$ $5.4\times {10}^{-6}=\frac{1}{2}\times 0.005\times v^2$

$\Rightarrow v=0.0465$ $m{s}^{-1}=4.65$ $m{s}^{-1}$