Question 756: Electrostats & 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 Nm2/C) _______.

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

Electrostats & Capacitance

Choose the correct answer: