Question

A point charge of 40 stat coulomb is placed 2 cm in front on an earthed metallic plane plate of large size. Then, the force of attraction on the point charge is
A. 100 dyne
B. 160 dyne
C. 1600 dyne
D. 400 dyne

Answer 1

Hint:-The charge is on the surface of the metallic conductor and due to the point charge 40 stat coulomb placed at 2cm far from plate, the same amount of charge will induce on plate. According to the coulomb’s law of force of attraction, it is directly proportional to the product of charges and inversely proportional to the square of distance between them.

Complete step-by-step solution:
Step 1:
Induction charging is a method used to charge an object without actually touching the object to any other charged object. An understanding of charging by induction requires an understanding of the nature of a conductor and an understanding of the polarization process.
Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.
It signifies the inverse square dependence of electric force. F=$k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ where, F is the electric force, ${q_1},{q_2}$ are the charges and r is the separation between them. Here, k is the coulomb constant and it is represented as$\dfrac{1}{{4\pi {e_0}}}$
Step 2:
Now coming to the question:
We are given that a point charge of 40 stat coulomb is placed 2 cm in front on an earthed metallic plane plate of large size.
Distance is 2 cm and charges and charges ${q_1},{q_2}$will be taken as 40 C because the same amount of charge is induced on the metallic plate.
See in the diagram:

The charges are equally placed on the metallic planes.

And then a charge of 40 stat coulomb placed 2 cm far from the plate. The charge which is positive is sent to the ground.
Then the induced charge on the metallic plate will be -40 stat C
Discussed above that the force of attraction is written as F=$k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ where, F is the electric force, ${q_1},{q_2}$ are the charges and r is the separation between them. Here, k is the coulomb constant and it is represented as$\dfrac{1}{{4\pi {e_0}}}$
Then, force of attraction is F=$\dfrac{1}{{4\pi {e_0}}}$×$\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
In CGS system the value of k(constant) is equal to 1
Substituting the values we get F=$1 \times \dfrac{{40 \times 40}}{{{2^2}}}$
This will give the force of attraction as 400 dynes
Hence option D is the correct answer.

Note:- The inside of a conductor cannot contain any net charge. Such charges would produce a field inside the conductor, and electrons would move and cancel out the field and neutralize the charge. Any excess charge on a conductor must therefore reside on the surface.