Suppose that electric potential due to a small charge configuration varies inversely with with square of distance from the charge distribution . Electric field intensity will vary inversely with what power of distance from the charge configuration in this case ?

Answer

To solve the problem, we need to analyze the relationship between electric potential (V) and electric field intensity (E) in the context of the given charge configuration.

### Step-by-Step Solution:

1. **Understand the relationship between electric potential and distance**:
We are given that the electric potential \( V \) due to a small charge configuration varies inversely with the square of the distance \( r \) from the charge distribution. This can be mathematically expressed as:
\[ V = \frac{k}{r^2} \]
where \( k \) is a constant.

2. **Relate electric field to electric potential**:
The electric field intensity \( E \) is related to the electric potential \( V \) by the formula:
\[ E = -\frac{dV}{dr} \]
This means that to find the electric field, we need to take the derivative of the potential with respect to distance.

3. **Differentiate the potential**:
We will differentiate \( V \) with respect to \( r \):
\[ V = k \cdot r^{-2} \]
Now, taking the derivative:
\[ \frac{dV}{dr} = k \cdot (-2) \cdot r^{-3} = -\frac{2k}{r^3} \]

4. **Calculate the electric field**:
Now substituting this derivative back into the equation for electric field:
\[ E = -\left(-\frac{2k}{r^3}\right) = \frac{2k}{r^3} \]
This shows that the electric field \( E \) is proportional to \( \frac{1}{r^3} \).

5. **Conclusion**:
From the above calculations, we conclude that the electric field intensity \( E \) varies inversely with the cube of the distance \( r \) from the charge configuration:
\[ E \propto \frac{1}{r^3} \]

### Final Answer:
The electric field intensity will vary inversely with the power of 3 of the distance from the charge configuration.

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