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. **Understanding the Relationship of Electric Potential**:
According to the question, 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:
\[ V \propto \frac{1}{R^2} \]
or
\[ V = \frac{K}{R^2} \]
where K is a constant.

2. **Finding the Electric Field Intensity**:
The electric field intensity (E) is related to the electric potential (V) by the formula:
\[ E = -\frac{dV}{dr} \]
We need to differentiate the expression for V with respect to R.

3. **Differentiating the Electric Potential**:
We have:
\[ V = \frac{K}{R^2} \]
To find E, we differentiate V:
\[ E = -\frac{d}{dr}\left(\frac{K}{R^2}\right) \]
Using the power rule for differentiation, we rewrite \( \frac{K}{R^2} \) as \( K \cdot R^{-2} \):
\[ E = -\frac{d}{dr}(K \cdot R^{-2}) = -K \cdot \left(-2R^{-3}\right) = \frac{2K}{R^3} \]

4. **Concluding the Relationship**:
From the differentiation, we find that:
\[ E = \frac{2K}{R^3} \]
This shows that the electric field intensity (E) varies inversely with the cube of the distance (R) from the charge configuration. Thus, we can conclude:
\[ E \propto \frac{1}{R^3} \]

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

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