The potential at a point distant x (mesured in `mum` ) due to some charges situated on the x-axis is given by `V (x)=(20)/(x^(2)-4)` V. The electric field at `x=4 mum` is given by

Options

A. `5/3 Vmu m^(-1)` and in positive x direction
B. `10/9 V mu m^(-1)` and in negative x direction
C. `10/9V mu m^(-1)` and in positive x direction
D. `5/3 V mu m^(-1)` and in negative x direction.

Explanation

To find the electric field at \( x = 4 \, \mu m \) given the potential \( V(x) = \frac{20}{x^2 - 4} \), we can use the relationship between electric field and electric potential. The electric field \( E \) is given by:

\[ E = -\frac{dV}{dx} \]

### Step 1: Differentiate the potential function

We start by differentiating \( V(x) \):

\[ V(x) = \frac{20}{x^2 - 4} \]

Using the quotient rule for differentiation, where if \( V = \frac{u}{v} \), then \( \frac{dV}{dx} = \frac{u'v - uv'}{v^2} \):

- Let \( u = 20 \) (constant), so \( u' = 0 \)
- Let \( v = x^2 - 4 \), so \( v' = 2x \)

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 2: Substitute into the electric field equation

Now, substituting this result into the equation for the electric field:

\[ E = -\frac{dV}{dx} = -\left(\frac{-40x}{(x^2 - 4)^2}\right) = \frac{40x}{(x^2 - 4)^2} \]

### Step 3: Evaluate the electric field at \( x = 4 \, \mu m \)

Now we substitute \( x = 4 \, \mu m \):

\[ E = \frac{40 \cdot 4}{(4^2 - 4)^2} \]

Calculating the denominator:

\[ 4^2 - 4 = 16 - 4 = 12 \]
\[ (12)^2 = 144 \]

Now substituting back into the equation for \( E \):

\[ E = \frac{160}{144} \]

### Step 4: Simplify the expression

Simplifying \( \frac{160}{144} \):

\[ E = \frac{10}{9} \, \text{V/}\mu m \]

### Final Answer

Thus, the electric field at \( x = 4 \, \mu m \) is:

\[ E = \frac{10}{9} \, \text{V/}\mu m \quad \text{in the positive x direction.} \]

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