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
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
Show 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.
Updated On: Jun 22, 2026
- \(-15\,\text{V}\)
- \(8\,\text{V}\)
- \(-10\,\text{V}\)
- \(-12.5\,\text{V}\)
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The Correct Option is C
Solution and Explanation
Step 1: Use the relation between electric field and potential difference.
The potential difference between two points is given by \[ V_B-V_A=-\int_A^B \vec{E}\cdot d\vec{r} \] Given electric field is \[ \vec{E}=5x\hat{i} \] This means the electric field has only \(x\)-component.
Step 2: Identify coordinates of points.
Point \(A\) lies on the \(y\)-axis at \(y=5\,\text{m}\). Therefore, \[ A=(0,5) \] Point \(B\) lies on the \(x\)-axis at \(x=2\,\text{m}\). Therefore, \[ B=(2,0) \] Since \(\vec{E}=5x\hat{i}\), only displacement in the \(x\)-direction contributes to the line integral.
Step 3: Calculate the potential difference.
We have \[ d\vec{r}=dx\hat{i}+dy\hat{j} \] So, \[ \vec{E}\cdot d\vec{r} = (5x\hat{i})\cdot(dx\hat{i}+dy\hat{j}) \] \[ \vec{E}\cdot d\vec{r}=5x\,dx \] Therefore, \[ V_B-V_A=-\int_0^2 5x\,dx \] \[ V_B-V_A=-5\int_0^2 x\,dx \] \[ V_B-V_A=-5\left[\frac{x^2}{2}\right]_0^2 \] \[ V_B-V_A=-5\left(\frac{4}{2}\right) \] \[ V_B-V_A=-10\,\text{V} \]
Step 4: Final conclusion.
Hence, \[ \boxed{-10\,\text{V}} \]
The potential difference between two points is given by \[ V_B-V_A=-\int_A^B \vec{E}\cdot d\vec{r} \] Given electric field is \[ \vec{E}=5x\hat{i} \] This means the electric field has only \(x\)-component.
Step 2: Identify coordinates of points.
Point \(A\) lies on the \(y\)-axis at \(y=5\,\text{m}\). Therefore, \[ A=(0,5) \] Point \(B\) lies on the \(x\)-axis at \(x=2\,\text{m}\). Therefore, \[ B=(2,0) \] Since \(\vec{E}=5x\hat{i}\), only displacement in the \(x\)-direction contributes to the line integral.
Step 3: Calculate the potential difference.
We have \[ d\vec{r}=dx\hat{i}+dy\hat{j} \] So, \[ \vec{E}\cdot d\vec{r} = (5x\hat{i})\cdot(dx\hat{i}+dy\hat{j}) \] \[ \vec{E}\cdot d\vec{r}=5x\,dx \] Therefore, \[ V_B-V_A=-\int_0^2 5x\,dx \] \[ V_B-V_A=-5\int_0^2 x\,dx \] \[ V_B-V_A=-5\left[\frac{x^2}{2}\right]_0^2 \] \[ V_B-V_A=-5\left(\frac{4}{2}\right) \] \[ V_B-V_A=-10\,\text{V} \]
Step 4: Final conclusion.
Hence, \[ \boxed{-10\,\text{V}} \]
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