The potential at a point x ( measured in `mu` m) due to some charges situated on the x-axis is given by
`V(x)=20//(x^2-4) vol t`

Options

A. `5//3" volt"//mum` and in the negative x-direction
B. `5//3" volt"//mum` and in the positive x-direction
C. `10//9" volt"//mum` and in the negative x-direction
D. `10//9" volt"//mum` and in the positive x-direction

Answer

To solve the problem, we need to find the electric field \( E \) at a point \( x = 4 \, \mu m \) using the given potential function \( V(x) = \frac{20}{x^2 - 4} \).

### Step 1: Understand the relationship between electric field and potential
The electric field \( E \) is related to the electric potential \( V \) by the formula:
\[ E = -\frac{dV}{dx} \]
This means we need to differentiate the potential function \( V(x) \) with respect to \( x \).

### Step 2: Differentiate the potential function
Given:
\[ V(x) = \frac{20}{x^2 - 4} \]
We will apply the quotient rule for differentiation. The quotient rule states that if \( V(x) = \frac{f(x)}{g(x)} \), then:
\[ \frac{dV}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]
Here, \( f(x) = 20 \) and \( g(x) = x^2 - 4 \).

Calculating the derivatives:
- \( f'(x) = 0 \) (since the derivative of a constant is zero)
- \( g'(x) = 2x \)

Now applying the quotient rule:
\[ \frac{dV}{dx} = \frac{0 \cdot (x^2 - 4) - 20 \cdot (2x)}{(x^2 - 4)^2} = \frac{-40x}{(x^2 - 4)^2} \]

### Step 3: Substitute \( x = 4 \, \mu m \) into the derivative
Now, we substitute \( x = 4 \, \mu m \) into the expression for \( \frac{dV}{dx} \):
\[ \frac{dV}{dx} = \frac{-40(4)}{(4^2 - 4)^2} = \frac{-160}{(16 - 4)^2} = \frac{-160}{12^2} = \frac{-160}{144} \]

### Step 4: Calculate the electric field \( E \)
Now we can find \( E \):
\[ E = -\frac{dV}{dx} = -\left(\frac{-160}{144}\right) = \frac{160}{144} = \frac{10}{9} \, \mu V/\mu m \]

### Step 5: Conclusion
Thus, the electric field at \( x = 4 \, \mu m \) is:
\[ E = \frac{10}{9} \, \mu V/\mu m \]