Electric potential at a point P, r distance away due to a point charge q kept at point A is V. If twice of this charge is distributed uniformly on the surface of a hollow sphere of radius 4r with centre at point A the potential at P now is

Options

A. V
B. `V//2`
C. `V//4`
D. `V//8`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial condition
The electric potential \( V \) at point \( P \) due to a point charge \( q \) located at point \( A \) and at a distance \( r \) is given by the formula:
\[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{r} \]

### Step 2: Determine the new charge and its distribution
The problem states that the charge is doubled, so the new charge \( Q \) is:
\[ Q = 2q \]
This charge is uniformly distributed over the surface of a hollow sphere with radius \( 4r \).

### Step 3: Calculate the potential at point \( P \) due to the hollow sphere
For a uniformly charged hollow sphere, the electric potential outside the sphere (at a distance greater than the radius of the sphere) is the same as if the entire charge were concentrated at the center of the sphere. Since point \( P \) is at a distance \( r \) from point \( A \) and the radius of the sphere is \( 4r \), point \( P \) is outside the sphere.

The potential \( V' \) at point \( P \) due to the hollow sphere can be calculated as:
\[ V' = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{d} \]
where \( d \) is the distance from the center of the sphere to point \( P \). Here, \( d = 4r \).

Substituting \( Q = 2q \) and \( d = 4r \):
\[ V' = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2q}{4r} \]
\[ V' = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{2r} \]

### Step 4: Relate the new potential to the initial potential
From the initial condition, we know:
\[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{r} \]

Now, we can express \( V' \) in terms of \( V \):
\[ V' = \frac{1}{2} \cdot \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{r} = \frac{1}{2} V \]

### Step 5: Conclusion
Thus, the potential at point \( P \) after distributing the charge on the hollow sphere is:
\[ V' = \frac{1}{2} V \]

### Final Answer
The potential at point \( P \) now is \( \frac{1}{2} V \).

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