An electric field $\stackrel{m \mathrm{E}}{\mathrm{E}}=4 x \hat{\mathrm{i}}-\left(\mathrm{y}^2+1\right) \hat{\mathrm{j}} \mathrm{N} / \mathrm{C}$ passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as $\varphi_{\mathrm{I}}$ and $\varphi_{\mathrm{II}}$ respectively. The difference between $\left(\varphi_1-\varphi_{\|}\right)$is (in $\left.\mathrm{Nm}^2 / \mathrm{C}\right)$ $\_\_\_\_$ .
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇