Question

How much greater is one microcoulomb compared to an electronic charge.
$
  {\text{A}}{.10^{13}}{\text{ times}} \\
  {\text{B}}{.10^{10}}{\text{ times}} \\
  {\text{C}}{.10^{11}}{\text{ times}} \\
  {\text{D}}{.10^6}{\text{ times}} \\
 $

Answer 1

Hint- The term Coulomb is the SI unit of the electric charge and it can be defined in terms of the ampere.

Complete step by step answer:
Electric charge is an intrinsic property of an elementary particle of matter which gives rise to electric force between various objects.
We can take down the values of 1 microcoulomb and 1 electric charge.
We have, 1 microcoulomb = ${10^{ - 6}}$coulomb
And 1 electric charge = $1.6 \times {10^{ - 19}}$coulomb.
We can see that; one electric charge is of the order of ${10^{ - 19}}$ and 1 micro coulomb is of the order of ${10^{ - 6}}$.
Therefore,
 $\dfrac{{{{10}^{ - 6}}}}{{{{10}^{ - 19}}}} = {10^{13}}$ times greater than an electric charge.
Hence, option (A) is the correct answer.

Additional information:
From the study of atomic structure, we know that an atom consists of a central part called nucleus and around the nucleus, there are a number of electrons revolving in a different path or orbit. The nucleus contains protons and neutrons. A proton is a positively charged particle while a neutron has no charge. Therefore, the nucleus of an atom gains a positive charge. An electron is a negatively charged particle having negative charge equal to the number of protons in an atom. Therefore, an atom is neutral as a whole; the negative charge on the electron cancels the positive charge on proton. This leads to the conclusion that, under ordinary conditions, a body is neutral, that is, it exhibits no charge.
According to Coulomb's law, the electrostatic force between two stationary point charges is directly proportional to the product of their magnitude and inversely proportional to the square of distance between their centres.
Let us consider the values and substitute in the below formula.

$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}} = 9 \times {10^9}{\text{N}}$.
Where,
$q_1, q_2$ are charges
$r$ is the distance

Note: The physical properties of matter that experience the force when placed in the electro-magnetic field is known as the electric charge. The electron charge is given as $1.602 \times {10^{ - 19}}C$.