Three metallic spheres of radii r,2r and 3r are taken.first sphere is intially given a charge q and other two are uncharged. third sphere is touched with first sphere then with second and finaaly removed. ratio of electric field intensity on surfaces 1 and 2 is?

Question

Answer

Answer

Answer

Given:
Three metallic spheres of radii r, 2r, and 3r.
First sphere is initially given a charge q.
The other two spheres are uncharged.

To find:
The ratio of the electric field intensity on surfaces 1 and 2.

Explanation:
Step 1: Charging the spheres
The first sphere is given a charge q.
The second and third spheres are uncharged initially.

Step 2: Touching the spheres
The third sphere is touched with the first sphere.
This results in the transfer of charge from the first sphere to the third sphere until both spheres reach the same potential.

Step 3: Touching the spheres again
The third sphere is then touched with the second sphere.
This results in the transfer of charge from the third sphere to the second sphere until both spheres reach the same potential.

Step 4: Removal of the third sphere
The third sphere is removed from the system.

Electric Field Intensity on Surface 1:
Since the first sphere is charged with charge q, it will create an electric field around it.
The electric field intensity on the surface of the first sphere will be given by the formula E = k*q/r^2, where k is the electrostatic constant.

Electric Field Intensity on Surface 2:
Since the second sphere is now charged due to the transfer of charge from the third sphere, it will also create an electric field around it.
The electric field intensity on the surface of the second sphere will be given by the formula E = k*q/4r^2, where q/4 is the charge acquired by the second sphere.

Ratio of Electric Field Intensities:
To find the ratio of the electric field intensities on the surfaces 1 and 2, we can divide the electric field intensity on surface 1 by the electric field intensity on surface 2.

E1/E2 = (k*q/r^2) / (k*q/4r^2)
= (4r^2)/(r^2)
= 4

Therefore, the ratio of the electric field intensity on surfaces 1 and 2 is 4.

Conclusion:
The ratio of the electric field intensity on surfaces 1 and 2 is 4.

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