A phserical shell of radius R and carrying charge Q. Now, if we decrease radius by (R/3) keeping the charge constant thena)The electrostatic energy will decrease by a factor 3b)The electrostatic energy will decrease by a factor 1.5c)The electrostatic energy will increase by a factor 3d)The electrostatic energywill be increase by a factor 1.5Correct answer is option 'D'. Can you explain this answer?

Question

Answer

Explanation:

To understand why the electrostatic energy will increase by a factor of 1.5 when the radius of the spherical shell is decreased by (R/3) while keeping the charge constant, we need to analyze the relationship between electrostatic energy and the radius of a charged sphere.

1. Electrostatic Energy:
The electrostatic energy of a charged sphere is given by the formula:

E = (3/5) * (k * Q^2) / R

where E is the electrostatic energy, k is the electrostatic constant, Q is the charge on the sphere, and R is the radius of the sphere.

2. Decreasing the Radius:
Now, let's consider the scenario where the radius of the spherical shell is decreased by (R/3) while keeping the charge constant. This means the new radius of the shell will be (2R/3).

3. Calculating the New Electrostatic Energy:
Using the formula for electrostatic energy, we can calculate the new electrostatic energy:

E' = (3/5) * (k * Q^2) / (2R/3)

Simplifying the expression:

E' = (9/10) * (k * Q^2) / R

4. Comparing the Original and New Energy:
To determine the factor by which the electrostatic energy changes, we can calculate the ratio of the new energy (E') to the original energy (E):

E' / E = [(9/10) * (k * Q^2) / R] / [(3/5) * (k * Q^2) / R]

Simplifying the expression:

E' / E = (9/10) / (3/5) = (9/10) * (5/3) = 3/2 = 1.5

Therefore, the electrostatic energy will increase by a factor of 1.5 when the radius of the spherical shell is decreased by (R/3) while keeping the charge constant.

Conclusion:
The correct answer is option 'D': The electrostatic energy will increase by a factor of 1.5.

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