Question

When a glass rod is rubbed with resin, a mass of $27.3\times {10}^{-14}g$ is transferred from glass to resin. Find the amount of charge acquired by the glass rod. What is the nature of the charge? (Charge of electron $=1.6\times {10}^{-19}C$, Mass of electron $=9.1\times {10}^{-31}kg$)

Answer 1

Step 1: Given the data and formula used:

Mass $M=$$27.3\times {10}^{-14}g$

Charge of electron $e=1.6\times {10}^{-19}C$

Mass of electron $m_e=9.1\times {10}^{-31}kg$

Mass $=$$27.3\times {10}^{-14}g$

We know that $1g={10}^{-3}kg$

Mass $=27.3\times {10}^{-14}\times {10}^{-3}kg$

$=27.3\times {10}^{-17}kg$

We know that,

$M=n\times m_e$

$n=\frac{M}{m_e}$

Where $n$ is the number of electrons, $m_e$ is the mass of the electron and $M$ is the mass of the charges that are transferred.

Step 2: Finding the amount of charge acquired by the glass rod:

Substituting the values

$$\begin{gathered} n=\frac{27.3\times {10}^{-17}}{9.1\times {10}^{-31}} \\ n=3\times {10}^{14} \end{gathered}$$

We know that, $Q=ne$

Substituting the values in the above formula

$$\begin{gathered} Q=3\times {10}^{14}\times 1.6\times {10}^{-19} \\ Q=Q=4.8\times {10}^{-5}C \\ Q=48\times {10}^{-6}C \end{gathered}$$

We know that $1\mu C={10}^{-6}C$ then,

$Q=48\mu C$

Hence, the amount of charge acquired by the glass rod is $48\mu C$ and the nature of the charge is positive.