Search any question & find its solution
Question:
Answered & Verified by Expert
Potential difference is given as $V(x)=-x^2 y$ volt. Find electric field at a point $(1,2)$ ?
Options:
Solution:
1552 Upvotes
Verified Answer
The correct answer is:
$4 \hat{\mathbf{i}}+\hat{\mathbf{j}} \mathrm{V} / \mathrm{m}$
Potential difference
$\begin{aligned}
V(x) & =-x^2 y V \\
E & =-\Delta V=-\left[\hat{i} \frac{\partial}{\partial x}\left(-x^2 y\right)+\hat{j} \frac{\partial}{\partial y}\left(-x^2 y\right)\right]
\end{aligned}$
$\begin{aligned} & =-\left[-2 x y \hat{i}-x^2 \hat{j}\right]=2 x y \hat{i}+x^2 \hat{j} \\ \text { E at }(1,2) & =2 \times 1 \times 2 \hat{i}+1^2 \hat{j} \\ & =4 \hat{i}+\hat{j} V / m\end{aligned}$
$\begin{aligned}
V(x) & =-x^2 y V \\
E & =-\Delta V=-\left[\hat{i} \frac{\partial}{\partial x}\left(-x^2 y\right)+\hat{j} \frac{\partial}{\partial y}\left(-x^2 y\right)\right]
\end{aligned}$
$\begin{aligned} & =-\left[-2 x y \hat{i}-x^2 \hat{j}\right]=2 x y \hat{i}+x^2 \hat{j} \\ \text { E at }(1,2) & =2 \times 1 \times 2 \hat{i}+1^2 \hat{j} \\ & =4 \hat{i}+\hat{j} V / m\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.