Question

A body is charged by rubbing it, then:
($q=$ net charge, $n_{p^{+}}=$ total number of protons after rubbing, $n_{e^{-}}=$ total number of electrons after rubbing)

• A. Its weight remains constant and charge on the body is given by $q=+\left(n_{p^{+}}\right)e$
• B. Its weight decreases slightly and charge on the body is given by $q=-\left(n_{e^{-}}\right)e$
• C. Its weight may increase slightly or may decrease slightly and charge on body is given by $q=(n_{p^{+}}-n_{e^{-}})e$
• D. Its weight increases slightly and charge on body is given by $q=+\left(n_{p^{+}}\right)e$

Answer 1

The correct option is C Its weight may increase slightly or may decrease slightly and charge on body is given by $q=(n_{p^{+}}-n_{e^{-}})e$

The gain and loss in weight of the body depends on the type of charge transfer during rubbing.
If a body has lost electrons during rubbing, then it will become positively charged and its weight will decrease.
Similarly, if a body has gained electrons during rubbing, it will become negatively charged and its weight will increase.
So, weight of body may decrease or may increase slightly.
In any of the case, net charge on body is given by,
$q=(n_{p^{+}}-n_{e^{-}})e$
$n_{p^{+}}=$ total number of protons after rubbing
$n_{e^{-}}=$ total number of electrons after rubbing

$\begin{array}{c}\text{Why this question ?}\\ \text{Bottom line: Presence of +ve and -ve charge on}\\ \text{body signifies loss and gain of electrons}\\ \text{by body, thus weight of body will change}\\ \text{slightly accordingly.}\end{array}$