Problem 87 - A Complete Resource Book in Physics for JEE Main

Question

A uniform electric field \(E=E_{0}(\hat{i}+\hat{j})\) exists in the region. The potential difference \(\left(V_{Q}-V_{P}\right)\) between point \(P(0,0)\) and \(Q(a, 0)\) is (A) \(-E_{0} a\) (B) \(E_{0} \sqrt{2} a\) (C) \(+E_{0} a\) (D) \(-E_{0} \sqrt{2} a\)

Step-by-Step Solution

Answer

The short answer is: \[V_{QP} = -E_{0}a\]

Step 1: Setup the integral for the potential difference

We need to calculate the integral \(V_{QP} = - \int_{P}^{Q} \vec E \cdot d \vec l\). The electric field is given as \(E = E_0(\hat{i} + \hat{j})\). The path we need to follow is from \(P(0, 0)\) to \(Q(a, 0)\). This path is along the x-axis. Now, since the path is along the x-axis, we can express the displacement vector \(d \vec l\) as \(dx \hat i\). Hence the dot product \(\vec E \cdot d \vec l = E_0(\hat{i} + \hat{j}) \cdot (dx\hat{i})= E_0 dx\). Now we can write the integral as: \[V_{QP}=-\int_{0}^{a} E_{0} dx\]

Step 2: Solving the integral

We can now proceed to evaluate the integral from Step 1: \[V_{QP} = - \int_{0}^{a} E_{0} dx\] Since \(E_0\) is a constant, we can take it outside the integral: \[V_{QP} = - E_{0} \int_{0}^{a} dx\] Now we can integrate: \[V_{QP}=-E_{0}\left[x\right]_{0}^{a}\] Evaluating the integral at the limits: \[V_{QP} = -E_{0}(a - 0)\] Thus, the potential difference is: \[V_{QP} = -E_{0}a\] So, the correct answer is: (A) \(-E_{0}a\)