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. (a) `(10//9) vol t//mu m` and in the `+ve` x direction
B. (b) `(5//3)vol t//mu m` and in the `-ve` x direction
C. (c) `(5//3) vol t//mu m` and in the `+ve` x direction
D. (d) `(10//9) vol t//mu m` and in the `-ve` x direction

Explanation

To find the electric field \( E \) at a point \( x \) due to the potential \( V(x) = \frac{20}{x^2 - 4} \) volts, we will follow these steps:

### Step 1: Understand the relationship between electric field and potential
The electric field \( E \) is related to the electric potential \( V \) by the equation:
\[ E = -\frac{dV}{dx} \]

### Step 2: Differentiate the potential function
We need to differentiate \( V(x) \) with respect to \( x \):
\[ V(x) = \frac{20}{x^2 - 4} \]
Using the quotient rule for differentiation, where if \( V = \frac{u}{v} \), then \( \frac{dV}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \):
- Here, \( u = 20 \) (a constant) and \( v = x^2 - 4 \).

Calculating the derivatives:
- \( \frac{du}{dx} = 0 \)
- \( \frac{dv}{dx} = 2x \)

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

### Step 3: Substitute into the electric field equation
Now substituting this result into the electric field equation:
\[ E = -\frac{dV}{dx} = -\left(\frac{-40x}{(x^2 - 4)^2}\right) = \frac{40x}{(x^2 - 4)^2} \]

### Step 4: Evaluate the electric field at \( x = 4 \mu m \)
Now we need to find the electric field at \( x = 4 \mu m \):
\[ E(4) = \frac{40 \times 4}{(4^2 - 4)^2} \]
Calculating the denominator:
\[ 4^2 - 4 = 16 - 4 = 12 \]
So,
\[ E(4) = \frac{40 \times 4}{12^2} = \frac{160}{144} \]

### Step 5: Simplify the result
Simplifying \( \frac{160}{144} \):
\[ E(4) = \frac{10}{9} \text{ (approximately 1.11)} \]

### Final Result
The electric field at \( x = 4 \mu m \) is:
\[ E(4) = \frac{10}{9} \, \text{N/m} \]

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