In Region of Electric field Given by `vec(E)=(Ax-B)hat(I)`. Where `A=20` unit and `B=10` unit. If Electric potential at `x=1m` is `v_(1)` and at `x=-5m` is `v_(2)`. Then `v_(1)-v_(2)` is equal to

Options

A. `150V`
B. `170V`
C. `140V`
D. `180V`

Answer

To find the difference in electric potential \( V_1 - V_2 \) given the electric field \( \vec{E} = (Ax - B) \hat{i} \), where \( A = 20 \) units and \( B = 10 \) units, we can follow these steps:

### Step 1: Understand the relationship between electric field and electric potential
The relationship between electric field \( \vec{E} \) and electric potential \( V \) is given by:
\[ \vec{E} = -\frac{dV}{dx} \]
This means that the change in electric potential can be found by integrating the electric field.

### Step 2: Set up the integral
We can express the change in potential \( V_1 - V_2 \) as:
\[ V_1 - V_2 = -\int_{x_2}^{x_1} \vec{E} \cdot d\vec{x} \]
In this case, \( x_1 = 1 \, m \) and \( x_2 = -5 \, m \).

### Step 3: Substitute the electric field into the integral
Substituting \( \vec{E} = (20x - 10) \hat{i} \) into the integral:
\[ V_1 - V_2 = -\int_{-5}^{1} (20x - 10) \, dx \]

### Step 4: Perform the integration
Now we can compute the integral:
\[ \int (20x - 10) \, dx = 10x^2 - 10x \]
Now we evaluate this from \( x = -5 \) to \( x = 1 \):
\[ V_1 - V_2 = -\left[ \left(10(1)^2 - 10(1)\right) - \left(10(-5)^2 - 10(-5)\right) \right] \]

### Step 5: Calculate the values
Calculating the first part:
\[ 10(1)^2 - 10(1) = 10 - 10 = 0 \]
Calculating the second part:
\[ 10(-5)^2 - 10(-5) = 10(25) + 50 = 250 + 50 = 300 \]
Now substituting back into the equation:
\[ V_1 - V_2 = -[0 - 300] = 300 \, V \]

### Step 6: Final result
Thus, the difference in electric potential is:
\[ V_1 - V_2 = 300 \, V \]