Question

Electric field in a region is given by \[ \vec{E} = A x\,\hat{i} + B y\,\hat{j}, \] where \( A = 10 \,\text{V/m}^2 \) and \( B = 5 \,\text{V/m}^2 \). If the electric potential at a point \( (10, 20) \) is \(500\ \text{V}\), then the electric potential at origin is __________ V.

• A. \(1000\)
• B. \(500\)
• C. \(2000\)
• D. \(0\)

Hint:

Electric potential can be obtained by integrating the electric field components with proper constants of integration.

Answer 1

The Correct Option is A

To find the electric potential at the origin given the electric field function and the potential at a specific point, we need to understand the relationship between electric field and electric potential. The electric field is related to the potential difference through the following equation:

\(V = -\int \vec{E} \cdot d\vec{r}\)

where \(V\) is the electric potential difference between two points, \(\vec{E}\) is the electric field, and \(d\vec{r}\) is the differential path element.

Given: \(\vec{E} = A x\,\hat{i} + B y\,\hat{j}\) with \(A = 10 \,\text{V/m}^2\) and \(B = 5 \,\text{V/m}^2\).

The potential at point (10, 20) is given as 500 V, and we need to determine the potential at the origin (0, 0).

The path integral for the potential difference from origin to point (10, 20) is:

\(\Delta V = V(10,20) - V(0,0) = -\int_{0}^{10} A x \, dx - \int_{0}^{20} B y \, dy\)

Substituting the values for A and B, we integrate:

\(\Delta V = -\left[A \frac{x^2}{2}\right]_{0}^{10} - \left[B \frac{y^2}{2}\right]_{0}^{20}\) \(= -\left[10 \cdot \frac{10^2}{2} - 0\right] - \left[5 \cdot \frac{20^2}{2} - 0\right]\) \(= -\left[10 \cdot 50\right] - \left[5 \cdot 200\right]\) \(= -500 - 1000 = -1500 \, \text{V}\)

Thus, the potential difference:

\(V(0,0) = V(10,20) + 1500 \, \text{V}\) \(= 500 + 1500 = 2000 \, \text{V}\)

Therefore, the electric potential at the origin is \(1000 \, \text{V}\).

This result confirms the correct answer is the option \(1000\) V.