Question

A neutral spherical copper particle has a radius of 10 nm (1nm=${10^{ - 9}}$m). It gets charged by applying the voltage slowly adding one electron at a time. Then the graph of the total charge on the particle vs. the applied voltage would look like:
A.
B.
C.
D.

Answer 1

In this question we will come across the concepts like the electric charge distribution and the potential of a spherical body. This question put light on these interesting topics. We will be learning all these concepts with all their related information in detail which will lead us in approaching our answer. Below here each topic is explained in a brief way.

Complete answer:
Electric field: An electric charge is surrounded by an electric field and applies a force on different charges present in the field. Which can be of attraction or repulsion. Electric fields are the invasion of electric charges and by time-changing magnetic fields.
Here we can put light on Gauss' law to understand the electric field of a charged sphere which tells us that the electric field environment on the outskirts of the sphere is the same as that of a point charge.
Potential: It always comes into account with the electric field. It is always attached and attributed to the electric field. It is donated by capital (v). Here we have talked about spherical potential so as the electric field is zero at centre so potential remains similar inside a conducting sphere.
For spherical surface
Potential = $V = \dfrac{{kQ}}{r}$
∵ Charge rises in distinct way which brings us to a result that, time charge is similar on sphere, the potential will remain same.
So, as we can see option (A) is right as it explains the condition in a proper way.

Hence, the correct option is A.

Note: In this problem we learned the concept of potential and distribution of charge in a sphere. Charge always tends to come to the surface if there is a presence of conductor as the conductor allows free movement of charge. The information we gained here we will be going to use in further questions also.