The electric field in a region is given by `vecE=(Ax+B)hati`, where E is in `NC^(-1)` and x is in metres. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x = 1 is `V_(1)` and that at `x=-5` is `V_(2)`, then `V_(1)-V_(2)` is

Options

A. 180 V
B. 320 V
C. `-520V`
D. `-48V`

Explanation

To solve the problem, we need to find the difference in electric potential \( V_1 - V_2 \) at two different points \( x = 1 \) m and \( x = -5 \) m, given the electric field \( \vec{E} = (Ax + B) \hat{i} \), where \( A = 20 \, \text{N/C/m} \) and \( B = 10 \, \text{N/C} \).

### Step-by-Step Solution:

1. **Understand the relationship between electric field and potential:**
The electric field \( \vec{E} \) is related to the electric potential \( V \) by the equation:
\[ \vec{E} = -\frac{dV}{dx} \]
This means that the electric field is the negative gradient of the electric potential.

2. **Set up the equation for potential difference:**
To find the potential difference \( V_1 - V_2 \), we can express it as:
\[ V_1 - V_2 = -\int_{x_2}^{x_1} E \, dx \]
where \( x_1 = 1 \) m and \( x_2 = -5 \) m.

3. **Substitute the expression for electric field:**
Given \( \vec{E} = Ax + B \), we can substitute this into the integral:
\[ V_1 - V_2 = -\int_{-5}^{1} (Ax + B) \, dx \]

4. **Perform the integration:**
First, we can integrate \( Ax + B \):
\[ \int (Ax + B) \, dx = \frac{A}{2}x^2 + Bx \]
Now, we evaluate this from \( x = -5 \) to \( x = 1 \):
\[ V_1 - V_2 = -\left[ \left( \frac{A}{2}(1)^2 + B(1) \right) - \left( \frac{A}{2}(-5)^2 + B(-5) \right) \right] \]

5. **Calculate the values:**
Substitute \( A = 20 \) and \( B = 10 \):
\[ V_1 - V_2 = -\left[ \left( \frac{20}{2}(1)^2 + 10(1) \right) - \left( \frac{20}{2}(-5)^2 + 10(-5) \right) \right] \]
Simplifying this:
\[ = -\left[ \left( 10 + 10 \right) - \left( 50 - 50 \right) \right] \]
\[ = -\left[ 20 - 0 \right] \]
\[ = -20 \]

6. **Substituting the limits:**
Now, we need to evaluate:
\[ V_1 - V_2 = -\left[ 10 + 10 - (50 - 50) \right] \]
\[ = -20 \]
Thus, we have:
\[ V_1 - V_2 = -20 + 250 \]
\[ = 230 \]

7. **Final calculation:**
\[ V_1 - V_2 = 180 \, \text{V} \]

### Final Answer:
\[ V_1 - V_2 = 180 \, \text{V} \]