Question

In a region electric field is given by $\vec{E} = 4x\,\hat{i}$ N/C. The potential difference between points A $(x = 1$ m$)$ and B $(x = 3$ m$)$, $(V_A - V_B)$ is:

• A. $-16$ V
• B. $16$ V
• C. $-8$ V
• D. $8$ V

Hint:

Integrate the field from x = 1 m to x = 3 m. Potential is higher at the point lying behind along the field direction.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The electric field is the negative slope of the potential, \(E = -\dfrac{dV}{dx}\).
So the potential difference between two points is the line integral of the field between them.
Here the field depends on \(x\), so we cannot just multiply \(E\) by the distance.

Step 2: Key Formula or Approach:
For a path from A to B along the field line:
\[ V_A - V_B = \int_{A}^{B} \vec{E}\cdot d\vec{x} \]
Point A is at \(x = 1\) m and point B is at \(x = 3\) m, and \(\vec{E} = 4x\,\hat{i}\) N/C.

Step 3: Detailed Explanation:
Put the field into the integral and use the limits 1 m to 3 m.
\[ V_A - V_B = \int_{1}^{3} 4x\,dx = 4\left[\frac{x^2}{2}\right]_{1}^{3} \]
\[ V_A - V_B = 2\,(3^2 - 1^2) = 2 \times 8 = 16\ \text{V} \]
The value is positive. The field points along \(+x\), and potential falls along the field. So A, which lies at smaller \(x\), is at the higher potential.

Step 4: Check the options:
Option (A) \(-16\) V has the right size but the wrong sign. It would mean B is at higher potential, which contradicts the direction of the field.
Option (C) \(-8\) V has both the wrong size and the wrong sign.
Option (D) \(8\) V comes from forgetting the factor 4 in the field. The integral of \(x\) gives \(x^2/2 = 4\), and the factor 4 is needed to reach 16.
Option (B) \(16\) V matches our result.

Final Answer:
The potential difference \(V_A - V_B\) is 16 V, which is option (B). \[ \boxed{16\ \text{V}} \]