The electric potential at a point `(x,0,0)` is given by `V=[(1000)/(x)+(1500)/(x^(2))+(500)/(x^(3))]` "then the electric field at" `x=1` m is (in volt//m)

Options

A. (a)`550 (hatj+hatk)Vm^(-1)`
B. (b)`5500 hatiVm^(-1)`
C. (c)`(5500)/(sqrt2)(hatj+hatk)Vm^(-1)`
D. (d)`(5500)/(sqrt2)(hati+hatk)Vm^(-1)`

Answer

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

### Step 1: Write down the expression for electric potential
The electric potential \( V \) is given by:
\[ V(x) = \frac{1000}{x} + \frac{1500}{x^2} + \frac{500}{x^3} \]

### Step 2: Differentiate the potential to find the electric field
The electric field \( E \) is related to the electric potential by the formula:
\[ E = -\frac{dV}{dx} \]
We need to differentiate \( V(x) \) with respect to \( x \).

### Step 3: Differentiate each term of \( V(x) \)
We differentiate \( V(x) \):
1. For the first term \( \frac{1000}{x} \):
\[ \frac{d}{dx}\left(\frac{1000}{x}\right) = -\frac{1000}{x^2} \]
2. For the second term \( \frac{1500}{x^2} \):
\[ \frac{d}{dx}\left(\frac{1500}{x^2}\right) = -\frac{3000}{x^3} \]
3. For the third term \( \frac{500}{x^3} \):
\[ \frac{d}{dx}\left(\frac{500}{x^3}\right) = -\frac{1500}{x^4} \]

Combining these results, we have:
\[ \frac{dV}{dx} = -\frac{1000}{x^2} - \frac{3000}{x^3} - \frac{1500}{x^4} \]

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

### Step 5: Calculate the electric field
Now we can find the electric field:
\[ E = -\frac{dV}{dx} = -(-5500) = 5500 \, \text{V/m} \]

### Final Answer
The electric field at \( x = 1 \) m is:
\[ E = 5500 \, \text{V/m} \]