Question

A glass rod is charged to +8.0 nC by rubbing.

a. Have electrons been removed from the rod or protons added?

b. How many electrons have been removed or protons added?

Short answer

a. The electrons have been removed from the rod.

b. $N=5\times {10}^{10}electrons.$are removed.

Step-by-step solution

Content Introduction

According to the principle of quantization Any body's charge is an integral multiple of the charge on the integral. Furthermore, when electrons are taken from a body, the body's negative charge is also lost. A body becomes positively charged as a result of this.

Explanation (Part a)

The glass rod has free electrons that are not bound to the nucleus and can travel about freely. Rubbing charges the glass rod by $+8nC$, causing it to become positively charged. The electrons are withdrawn from the glass rod, and this is the only way to make the glass rod positively charged when protons cannot be added.

Explanation (Part b)

The charge is of electrons and protons like mass.

Here, $Q$is the charge and for glass rod its charge equals the number of carrier N times the charges of the carrier, which $-e$for electrons whereas, $+e$for protons. Therefore,

$Q=Ne$

where, $e=-1.6\times {10}^{-19}C$, $Q=-8nC$as it represents the number of charges that remove to make the glass rod$+8nC$

$$\begin{gathered} N=\frac{Q}{e} \\ N=\frac{-8\times {10}^{-9}C}{1.6\times {10}^{-19}C} \\ N=5\times {10}^{10}electrons \end{gathered}$$