A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity 20000 V / m. If mass of the particle is 9.6 × 10-16 kg, the charge on it and excess number of electrons on the-particle respectively are ( g =10 m / s 2)

Question

A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity 20000 V / m. If mass of the particle is 9.6 × 10-16 kg, the charge on it and excess number of electrons on the-particle respectively are ( g =10 m / s 2) (A) 4.8 × 10-19 C, 3 (B) 5.8 × 10-19 C, 4 (C) 3.8 × 10-19 C, 2 (D) 2.8 × 10-19 C, 1

Tardigrade Admin · Accepted Answer

Charge on the particle is given by q E =m g or q=(m g/E) =(9.6 × 10-16 × 10/20000)=4.8 × 10-19 When there is an excess of n electrons on the particle, then q=n e So, n=(q/e)=(4.8 × 10-19/1.6 × 10-19)=3

Q. A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20000 V / m$ . If mass of the particle is $9.6 \times 1 0^{- 16} k g$ , the charge on it and excess number of electrons on the-particle respectively are $(g = 10 m / s^{2})$

$4.8 \times 1 0^{- 19} C, 3$

$5.8 \times 1 0^{- 19} C, 4$

$3.8 \times 1 0^{- 19} C, 2$

$2.8 \times 1 0^{- 19} C, 1$

Charge on the particle is given by
$qE = m g$ or $q = \frac{m g}{E}$
$= \frac{9.6 \times 1 0 ^{- 16} \times 10}{20000} = 4.8 \times 1 0^{- 19}$
When there is an excess of $n$ electrons on the particle, then
$q = n e$
So, $n = \frac{q}{e} = \frac{4.8 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 3$

Q. A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity 20000V/m. If mass of the particle is 9.6×10−16kg, the charge on it and excess number of electrons on the-particle respectively are (g=10m/s2)

4.8×10−19C,3

5.8×10−19C,4

3.8×10−19C,2

2.8×10−19C,1