1
JEE Main 2021 (Online) 25th February Morning Shift
Numerical
+4
-1
The electric field in a region is given by $$\overrightarrow E = \left( {{3 \over 5}{E_0}\widehat i + {4 \over 5}{E_0}\widehat j} \right){N \over C}$$. The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y $$-$$ z plane) to that of the surface of area 0.3 m2 (parallel to x $$-$$ z plane) is a : b, where a = __________ [Here $${\widehat i}$$, $${\widehat j}$$ and $${\widehat k}$$ are unit vectors along x, y and z-axes respectively.]
2
JEE Main 2021 (Online) 25th February Morning Shift
Numerical
+4
-1
512 identical drops of mercury are charged to a potential of 2V each. The drops are joined to form a single drop. The potential of this drop is ________ V.
3
JEE Main 2021 (Online) 24th February Evening Shift
Numerical
+4
-1
A point charge of +12$$\mu$$C is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be __________ $$\times$$ 103 Nm2/C.

4
JEE Main 2020 (Online) 9th January Evening Slot
Numerical
+4
-0
An electric field $$\overrightarrow E = 4x\widehat i - \left( {{y^2} + 1} \right)\widehat j$$ N/C
passes through the box shown in figure. The
flux of the electric field through surfaces ABCD
and BCGF are marked as $${\phi _I}$$ and $${\phi _{II}}$$
respectively. The difference between $$\left( {{\phi _I} - {\phi _{II}}} \right)$$ is (in Nm2/C) _______.