Question 691: Electrostatics & Capacitance

Question

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 Nm
2
/C) _______.
![](https://imagex.cdn.examgoal.net/7k9k2k5le8nf1/86487429-85d9-426e-9171-39f5b1de394e/0f3863d0-3aef-11ea-92dd-4b9c3519996f/file-7k9k2k5le8nf2-430w.jpg)

Accepted Answer

Flux via ABCD
$\phi $
I
= $\int {\overrightarrow E } .d\overrightarrow A $ = 0
Flux via EFGH
$\phi $
II
= $\int {\overrightarrow E } .d\overrightarrow A $
= [$4x\widehat i - \left( {{y^2} + 1} ight)\widehat j$].4$\widehat i$
= 16x = 16 $ 	imes $ 3 = 48
${\phi _I} - {\phi _{II}}$ = 0 - 48 = -48  Nm
2
/C

Electrostatics & Capacitance

Integer / Numeric Answer Type