If the electric field is given by `vec(E ) = ((100)/(x^(2)))i` the potential difference between points x=10m and x=20m in volts is

Options

A. `-5`
B. 5
C. `-10`
D. 10

Explanation

To find the potential difference between the points \( x = 10 \, \text{m} \) and \( x = 20 \, \text{m} \) given the electric field \( \vec{E} = \frac{100}{x^2} \hat{i} \), we can follow these steps:

### Step 1: Understand the relationship between electric field and potential
The electric field \( \vec{E} \) is related to the electric potential \( V \) by the equation:
\[ \vec{E} = -\nabla V \]
In one dimension (along the x-axis), this simplifies to:
\[ E = -\frac{dV}{dx} \]

### Step 2: Set up the integral for potential difference
The potential difference \( V_{AB} \) between two points \( A \) and \( B \) can be calculated using the integral of the electric field:
\[ V_{AB} = V_B - V_A = -\int_{x_A}^{x_B} E \, dx \]
In our case, \( A \) is at \( x = 10 \, \text{m} \) and \( B \) is at \( x = 20 \, \text{m} \).

### Step 3: Substitute the electric field into the integral
Substituting \( E = \frac{100}{x^2} \):
\[ V_{20} - V_{10} = -\int_{10}^{20} \frac{100}{x^2} \, dx \]

### Step 4: Calculate the integral
The integral can be computed as follows:
\[ -\int_{10}^{20} \frac{100}{x^2} \, dx = -100 \left[ -\frac{1}{x} \right]_{10}^{20} \]
Calculating the limits:
\[ = -100 \left( -\frac{1}{20} + \frac{1}{10} \right) \]
\[ = -100 \left( -\frac{1}{20} + \frac{2}{20} \right) = -100 \left( \frac{1}{20} \right) \]
\[ = -5 \, \text{volts} \]

### Step 5: Conclusion
The potential difference between the points \( x = 10 \, \text{m} \) and \( x = 20 \, \text{m} \) is:
\[ V_{20} - V_{10} = -5 \, \text{volts} \]