The electric potential at a point `(x,0,0)` is given by `V=[(1000)/(x)+(1500)/(x)+(500)/(x^(3))]` "then the electric field at" `x=1` m is `(in volt//m)`

Options

A. `-5500hat(i)`
B. `5500hat(i)`
C. `sqrt(5500)hat(i)`
D. zero

Explanation

To find the electric field at the point \( x = 1 \) m from the given electric potential \( V(x) = \frac{1000}{x} + \frac{1500}{x^2} + \frac{500}{x^3} \), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Electric Field Formula**:
The electric field \( E \) is related to the electric potential \( V \) by the formula:
\[ E = -\frac{dV}{dx} \]

2. **Differentiate the Electric Potential**:
We need to differentiate the given potential \( V(x) \):
\[ V(x) = \frac{1000}{x} + \frac{1500}{x^2} + \frac{500}{x^3} \]
We will differentiate each term separately:
- The derivative of \( \frac{1000}{x} \) is \( -\frac{1000}{x^2} \)
- The derivative of \( \frac{1500}{x^2} \) is \( -\frac{3000}{x^3} \)
- The derivative of \( \frac{500}{x^3} \) is \( -\frac{1500}{x^4} \)

Therefore, we can write:
\[ \frac{dV}{dx} = -\frac{1000}{x^2} - \frac{3000}{x^3} - \frac{1500}{x^4} \]

3. **Substitute \( x = 1 \) m**:
Now we will substitute \( x = 1 \) into the derivative:
\[ \frac{dV}{dx} \bigg|_{x=1} = -\left( \frac{1000}{1^2} + \frac{3000}{1^3} + \frac{1500}{1^4} \right) \]
Simplifying this gives:
\[ \frac{dV}{dx} \bigg|_{x=1} = -\left( 1000 + 3000 + 1500 \right) = -5500 \]

4. **Calculate the Electric Field**:
Now we can find the electric field:
\[ E = -\frac{dV}{dx} = -(-5500) = 5500 \, \text{V/m} \]

5. **Final Result**:
The electric field at \( x = 1 \) m is:
\[ E = 5500 \, \hat{i} \, \text{V/m} \]