The electric potential `V` is givne as a function of distance `x` (metre) by `V = (5 x^(2) + 10x - 9)` volt. Value of electric field at `x = 1 `is

Options

A. `-20 V//m`
B. `6 V//m`
C. `11 V//m`
D. `-23 V//m`

Answer

To find the electric field \( E \) at \( x = 1 \) given the electric potential \( V \) as a function of distance \( x \), we will follow these steps:

### Step 1: Write down the expression for electric potential
The electric potential \( V \) is given by:
\[ V = 5x^2 + 10x - 9 \]

### Step 2: Differentiate the potential with respect to \( x \)
To find the electric field \( E \), we use the relation:
\[ E = -\frac{dV}{dx} \]
Now, we need to differentiate \( V \) with respect to \( x \):
\[ \frac{dV}{dx} = \frac{d}{dx}(5x^2 + 10x - 9) \]
Using the power rule of differentiation:
\[ \frac{dV}{dx} = 5 \cdot 2x + 10 \cdot 1 - 0 = 10x + 10 \]

### Step 3: Substitute \( x = 1 \) into the derivative
Now we substitute \( x = 1 \) into the expression we found for \( \frac{dV}{dx} \):
\[ \frac{dV}{dx} \bigg|_{x=1} = 10(1) + 10 = 10 + 10 = 20 \]

### Step 4: Calculate the electric field
Now, we can find the electric field \( E \):
\[ E = -\frac{dV}{dx} = -20 \, \text{V/m} \]

### Final Answer
The value of the electric field at \( x = 1 \) is:
\[ E = -20 \, \text{V/m} \]

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