Assume that an electric field `vecE=30x^2hatj` exists in space. Then the potential difference `V_A-V_O`, where `V_O` is the potential at the origin and `V_A` the potential at `x=2m` is:

Options

A. (a) `120J//C`
B. (b) `-120J//C`
C. (c) `-80J//C`
D. (d) `80J//C`

Explanation

To find the potential difference \( V_A - V_O \) where \( V_O \) is the potential at the origin and \( V_A \) is the potential at \( x = 2 \, \text{m} \), we start with the given electric field:

\[ \vec{E} = 30x^2 \hat{j} \]

### Step 1: Relate Electric Field to Potential
The relationship between electric field \( \vec{E} \) and electric potential \( V \) is given by:

\[ \vec{E} = -\frac{dV}{dx} \]

### Step 2: Set Up the Integral
To find the potential difference \( V_A - V_O \), we can rearrange the equation:

\[ V_A - V_O = -\int_{0}^{2} \vec{E} \cdot d\vec{x} \]

Since the electric field is in the \( \hat{j} \) direction, we need to consider the \( y \) component of the field. However, since the field is only in the \( y \) direction and we are moving in the \( x \) direction, we can express this as:

\[ V_A - V_O = -\int_{0}^{2} 30x^2 \, dx \]

### Step 3: Calculate the Integral
Now we compute the integral:

\[ V_A - V_O = -30 \int_{0}^{2} x^2 \, dx \]

Calculating the integral:

\[ \int x^2 \, dx = \frac{x^3}{3} \]

So we evaluate it from \( 0 \) to \( 2 \):

\[ \int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \]

### Step 4: Substitute Back into the Equation
Now substituting back into the equation for potential difference:

\[ V_A - V_O = -30 \left( \frac{8}{3} \right) \]

Calculating this gives:

\[ V_A - V_O = -80 \]

### Final Result
Thus, the potential difference \( V_A - V_O \) is:

\[ V_A - V_O = -80 \, \text{J/C} \quad \text{or} \quad -80 \, \text{V} \]