The electric field in a region is given by `oversettoE=(Ax+B)hati` Where E is in `NC^(-)` and x is in metres. The values of constants are `A=20SI` Unit and `B==10SI` 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. `320V`
B. `-48V`
C. `180V`
D. `-520V`

Answer

To solve the problem, we need to find the potential difference \( V_1 - V_2 \) given the electric field \( \overset{\rightarrow}{E} = (Ax + B) \hat{i} \), 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 \( E \) is related to the electric potential \( V \) by the equation:
\[ E = -\frac{dV}{dx} \]
This implies that the change in potential \( dV \) can be expressed as:
\[ dV = -E \, dx \]

2. **Substitute the expression for electric field:**
Given \( E = Ax + B \), we can write:
\[ dV = -(Ax + B) \, dx \]

3. **Integrate to find the potential difference:**
We want to find the potential difference between \( x = 1 \) and \( x = -5 \):
\[ V_1 - V_2 = -\int_{-5}^{1} (Ax + B) \, dx \]

4. **Set up the integral:**
Substitute \( A = 20 \) and \( B = 10 \):
\[ V_1 - V_2 = -\int_{-5}^{1} (20x + 10) \, dx \]

5. **Calculate the integral:**
First, compute the integral:
\[ \int (20x + 10) \, dx = 10x^2 + 10x \]
Now evaluate this from \( -5 \) to \( 1 \):
\[ V_1 - V_2 = -\left[ (10(1)^2 + 10(1)) - (10(-5)^2 + 10(-5)) \right] \]
Simplifying this gives:
\[ = -\left[ (10 + 10) - (10 \cdot 25 - 50) \right] \]
\[ = -\left[ 20 - (250 - 50) \right] \]
\[ = -\left[ 20 - 200 \right] \]
\[ = -(-180) = 180 \]

6. **Final Result:**
Thus, the potential difference \( V_1 - V_2 \) is:
\[ V_1 - V_2 = 180 \, \text{V} \]