JEE PYQ: Electrostatics Question 46
Question 46 - 2019 (10 Apr Shift 2)
In free space, a particle A of charge $1;\mu$C is held fixed at a point P. Another particle B of the same charge and mass $4;\mu$g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is:
$\left[\text{Take } \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9;\text{Nm}^2\text{C}^{-2}\right]$
(1) 1.0 m/s
(2) $3.0 \times 10^4$ m/s
(3) $2.0 \times 10^3$ m/s
(4) $1.5 \times 10^2$ m/s
Show Answer
Answer: (3)
Solution
$\frac{1}{2}mv^2 = kq_1 q_2\left(\frac{1}{r_1} - \frac{1}{r_2}\right) = 9 \times 10^9 \times (10^{-6})^2 \times \left(\frac{1}{10^{-3}} - \frac{1}{9 \times 10^{-3}}\right) = 9 \times 10^{-3} \times \frac{8}{9} = 8 \times 10^{-3}$. $v^2 = \frac{2 \times 8 \times 10^{-3}}{4 \times 10^{-9}} = 4 \times 10^6$. $v = 2 \times 10^3$ m/s.
BETA
