JEE Physics Electric Charges And Fields MCQs Set 07

Answer: (a) 1.76 mm

Explanation:
1. Time in the field \(t=\frac{0.10}{4\times10^7}=2.5\times10^{-9}\) s
2. Acceleration \(a=\frac{eE}{m}=1.76\times10^{11}\times3200=5.63\times10^{14}\) m/s\(^2\)
3. \(y=\tfrac12at^2=\tfrac12\times5.63\times10^{14}\times(2.5\times10^{-9})^2=1.76\times10^{-3}\) m = 1.76 mm

Teacher's Note:
1. Treat the motion like a horizontally thrown projectile: uniform speed along the plates and uniform acceleration across them.

Answer: (d) All of the above

Explanation:
1. The field between the plates is \(E=\sigma/\varepsilon_0\)
2. Along AB the force and displacement are parallel: \(W=q_0Ex=q_0\sigma x/\varepsilon_0\)
3. Along BC the force is perpendicular to the displacement, so the work is zero
4. The electrostatic field is conservative, so the work over a closed path is zero; all three statements are true

Teacher's Note:
1. Work along a path perpendicular to the field is zero and the net work over any closed loop in an electrostatic field is zero.

Answer: (c) parabola

Explanation:
1. The particle moves with constant velocity along the plates
2. Across the plates it has constant acceleration \(qE/m\)
3. \(y=\frac{qE}{2mu^2}x^2\) which is a parabola

Teacher's Note:
1. Uniform speed in one direction plus uniform acceleration in the perpendicular direction always gives a parabola.

Answer: (b) 90 m/s

Explanation:
1. Energy conservation: \(\tfrac12mv^2=kQq\left(\frac1{r_1}-\frac1{r_2}\right)\)
2. \(=9\times10^9\times10^{-3}\times10^{-6}\left(1-\tfrac1{10}\right)=8.1\) J
3. \(v=\sqrt{\frac{2\times8.1}{2\times10^{-3}}}=90\) m/s

Teacher's Note:
1. Convert grams to kilograms and micro and milli coulombs to coulombs before substituting. The loss of electric potential energy becomes kinetic energy.