JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Electrostatics

An electric field

E = 4 x \hat{i} - (y^{2} + 1) \hat{j}

N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as

ϕ_{I}

and

ϕ_{I I}

respectively. The difference between

(ϕ_{I} - ϕ_{I I})

is (in

Nm^{2} / C

) ____.

Answer: -48

Solution

Through these two faces only the

x

-component of the field,

E_{x} = 4 x

, can carry flux, because the area vectors of ABCD and BCGF point along the

x

-axis while the

- (y^{2} + 1)

component runs parallel to them and threads nothing. Over a face normal to the

x

-axis the coordinate

x

is constant, so the flux through it is simply

E_{x}

times that face's area. Evaluating

ϕ_{I}

for ABCD and

ϕ_{I I}

for BCGF at their positions set by the figure and subtracting gives

ϕ_{I} - ϕ_{I I} = - 48 Nm^{2} / C

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Integer answer