MCQ
An electron having charge $' e\ '$ and mass $' m\ '$ is moving in a uniform electric field $E .$ Its acceleration will be
- A$\frac{ e ^2}{m}$
- B$\frac{ E ^2 e }{ m }$
- ✓$\frac{ eE }{ m }$
- D$\frac{m E}{e}$
✓
Answer
Correct option: C.
$\frac{ eE }{ m }$
$\frac{ eE }{ m }$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
For a thermocouple, the temperature of inversion is that temperature at which thermo e.m.f. is(a) Zero (b) Maximum
(c) Minimum(d) None of the above
(c) Minimum(d) None of the above
The intensity of magnetic field is H and moment of magnet is M. The maximum potential energy is(a) MH(b) 2 MH(c) 3 MH (d) 4 MH
Two identical short bar magnets, each having magnetic moment of $10 \ \mathrm{Am}^2,$ are arranged such that their axial lines are perpendicular to each other and their centres be along the same straight line in a horizontal plane. If the distance between their centres is $0.2 m ,$ the resultant magnetic induction at a point midway between them is $(\left.\mu_0=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}\right)$
→Period of revolution increases in the order of (a) Saturn, Uranus, Venus(b) Mars, Saturn, Pluto(c) Mercury, Neptune, Mars(d) Mars, Jupiter, Venus
The two bulbs as in the above question are connected in series to a 200 volt line. Then(a) The potential drop across the two bulbs is the same(b) The potential drop across the 40 watt bulb is greater than the potential drop across the 100 watt bulb(c) The potential drop across the 100 W bulb is greater than the potential drop across the 40 W bulb(d) The potential drop across both the bulb is 200 volt
The $K_\alpha\ X-$ray emission line of tungsten occurs at $λ = 0.021\ nm.$ The energy difference between $K$ and $L$ levels in this atom is about
→When the light source is approaching, the Doppler effect for light is known as:
A man can see only between 75 cm and 200 cm. The power of lens to correct the near point will be(a) + 8/3 D(b) + 3 D(c) – 3 D (d) – 8/3 D
Shown below is a distribution of charges. The flux of electric field due to these charges through the surface $S$ is
→Cathode rays travelling from east to west enter into region of electric field directed towards north to south in the plane of paper. The deflection of cathode rays is towards(a) East (b) South(c) West (d) North