Question

Assume that an electric field $\vec{E}=30 x^{2} \hat{i}$ 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 $x=2 \,m$ is

• A. 120 J
• B. -120 J
• C. -80 J
• D. 80 J

Answer 1

The Correct Option is C

To find the potential difference \( V_{A} - V_{O'} \) where \( \vec{E} = 30 x^2 \hat{i} \) is given as the electric field, we start with the following relationship:

The potential difference between two points in an electric field is given by:

\[ V_{A} - V_{O} = -\int_{O}^{A} \vec{E} \cdot d\vec{r} \]

Here, \( \vec{E} \cdot d\vec{r} = E_x \, dx \) because the field is in the \( x \)-direction only. Hence, the integral becomes:

\[ V_{A} - V_{O} = -\int_{0}^{2} 30x^2 \, dx \]

Calculating this integral:

• First, integrate \( 30x^2 \):
• \[ \int 30x^2 \, dx = 30 \left( \frac{x^3}{3} \right) = 10x^3 \]

Now, evaluate this from 0 to 2:

\[ \left. 10x^3 \right|_{0}^{2} = 10(2^3) - 10(0^3) = 80 \]

Thus, the potential difference is:

\[ V_{A} - V_{O} = -80 \, \text{J} \]

Therefore, the potential difference \( V_{A} - V_{O} \) is \(-80 \, \text{J}\), which means the correct answer is -80 J.