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

Correct answer: A. 4.8 × 10^ -19 C, 3 Charge on the particle is given by aligned q E & =m g or q=(m g)/(E) \& = 9.6 × 10^ -16 × 10 20000 =4.8 × 10^ -19 aligned 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

JIPMER 2004

Exam: JIPMER
Year: 2004
Subject: Physics
Chapter: Electrostatics

JIPMER 2004 · Physics · Electrostatics

A charged particle is suspended in equilibrium in a uniform...

A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity

20000 \mathrm{~V} / \mathrm{m}

. If mass of the particle is

9.6 \times 10^{-16} \mathrm{~kg}

, the charge on it and excess number of electrons on the-particle respectively are

\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)

Options

A $4.8 \times 10^{-19} \mathrm{C}, 3$
B $5.8 \times 10^{-19} \mathrm{C}, 4$
C $3.8 \times 10^{-19} \mathrm{C}, 2$
D $28 \times 10^{-19} \mathrm{C}, 1$

Correct answer

A. $4.8 \times 10^{-19} \mathrm{C}, 3$

Step-by-step solution

Charge on the particle is given by

\begin{aligned}q E & =m g \text{ or } q=\frac{m g}{E} \& =\frac{9.6 \times 10^{-16} \times 10}{20000}=4.8 \times 10^{-19} \end{aligned}

When there is an excess of

n

electrons on the particle, then

q=n e

So,

n=\frac{q}{e}=\frac{4.8 \times 10^{-19}}{1.6 \times 10^{-19}}=3

Practice Electrostatics on Quantrex Academy →

More from Electrostatics