Question

The electric field in a region of space is given as \(\vec{E}=(5\,\text{N C}^{-1})x\hat{i}\). Consider point \(A\) on the \(y\)-axis at \(y=5\,\text{m}\) and point \(B\) on the \(x\)-axis at \(x=2\,\text{m}\). If the potentials at points \(A\) and \(B\) are \(V_A\) and \(V_B\) respectively, then \((V_B-V_A)\) is

• A. \(-15\,\text{V}\)
• B. \(8\,\text{V}\)
• C. \(-10\,\text{V}\)
• D. \(-12.5\,\text{V}\)

Hint:

Potential difference is calculated using \[ V_B-V_A=-\int_A^B \vec{E}\cdot d\vec{r} \] If the electric field has only \(x\)-component, then only \(dx\) contributes to the integral.

Answer 1

The Correct Option is C

Step 1: Write the relation between electric field and potential difference.
The potential difference between two points is given by: \[ V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{r} \] The electric field is $\vec{E} = 5x\,\hat{i}\,\text{N/C}$, which has only an $x$-component.
Step 2: Identify the coordinates of points A and B.
Point $A$ is on the $y$-axis at $y = 5\,\text{m}$, so $A = (0, 5, 0)$. Point $B$ is on the $x$-axis at $x = 2\,\text{m}$, so $B = (2, 0, 0)$.
Step 3: Choose a convenient integration path.
Since $\vec{E}$ is conservative, we can choose any path from $A$ to $B$. A convenient path is: first go from $A = (0, 5)$ to $O = (0, 0)$ along the $y$-axis, then from $O = (0, 0)$ to $B = (2, 0)$ along the $x$-axis.
Step 4: Calculate the potential difference along the first segment (A to O).
Along the $y$-axis, $x = 0$, so $\vec{E} = 0$ everywhere on this segment. Therefore: \[ V_O - V_A = -\int_A^O \vec{E} \cdot d\vec{r} = 0 \] So $V_O = V_A$.
Step 5: Calculate the potential difference along the second segment (O to B).
Along the $x$-axis, $d\vec{r} = dx\,\hat{i}$ and $\vec{E} = 5x\,\hat{i}$: \[ V_B - V_O = -\int_0^2 5x\,dx = -5\left[\frac{x^2}{2}\right]_0^2 = -5 \times \frac{4}{2} = -10\,\text{V} \]
Step 6: Combine the results to find $V_B - V_A$.
\[ V_B - V_A = (V_B - V_O) + (V_O - V_A) = -10 + 0 = -10\,\text{V} \] \[ \boxed{V_B - V_A = -10\,\text{V}} \]