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Electrostatics question

2020 · 9 Jan · Shift 2 · Q42
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Electrostatics question

2020 · 9 Jan · Shift 2 · Q42

JEE MainPhysicsElectrostaticsNumerical+4 / −1
An electric field E→=4xi^−(y2+1)j^\overrightarrow E = 4x\widehat i - \left( {{y^2} + 1} \right)\widehat jE=4xi−(y2+1)j​ N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as ϕI{\phi _I}ϕI​ and ϕII{\phi _{II}}ϕII​ respectively. The difference between (ϕI−ϕII)\left( {{\phi _I} - {\phi _{II}}} \right)(ϕI​−ϕII​) is (in Nm2/C) ‾\underline{\hspace{2cm}}​. JEE Main 2020 (Online) 9th January Evening Slot Physics - Electrostatics Question 176 English
Numerical answer
View written solutionFree

Correct answer: -48

  1. Interpret the field and the required faces

The electric field is

E⃗=4x i^−(y2+1) j^.\vec E = 4x\,\hat i - (y^2+1)\,\hat j.E=4xi^−(y2+1)j^​.

We need the flux through two faces of the rectangular box:

  • surface ABCD: this is a face perpendicular to the xxx-axis,
  • surface BCGF: this is a face perpendicular to the yyy-axis.

From the standard cuboid labeling in the figure, the box dimensions are:

  • along xxx: from x=0x=0x=0 to x=4x=4x=4,
  • along yyy: from y=0y=0y=0 to y=2y=2y=2,
  • along zzz: from z=0z=0z=0 to z=3z=3z=3.

So each relevant face has area:

  • face perpendicular to xxx: area =(Δy)(Δz)=2×3=6= (\Delta y)(\Delta z)=2\times 3=6=(Δy)(Δz)=2×3=6,
  • face perpendicular to yyy: area =(Δx)(Δz)=4×3=12= (\Delta x)(\Delta z)=4\times 3=12=(Δx)(Δz)=4×3=12.
  1. Flux through surface ABCDABCDABCD

Surface ABCDABCDABCD lies at constant x=0x=0x=0, and its outward area vector is along −i^-\hat i−i^.

At x=0x=0x=0,

Ex=4x=0.E_x = 4x = 0.Ex​=4x=0.

Hence

ϕI=∬E⃗⋅dA⃗=0.\phi_I = \iint \vec E\cdot d\vec A = 0.ϕI​=∬E⋅dA=0.
  1. Flux through surface BCGFBCGFBCGF

Surface BCGFBCGFBCGF lies at constant y=2y=2y=2, and its outward area vector is along +j^+\hat j+j^​.

The yyy-component of the field is

Ey=−(y2+1).E_y = -(y^2+1).Ey​=−(y2+1).

At y=2y=2y=2,

Ey=−(22+1)=−(4+1)=−5 N/C.E_y = -(2^2+1)=-(4+1)=-5\ \text{N/C}.Ey​=−(22+1)=−(4+1)=−5 N/C.

Area of this face:

A=(4)(3)=12.A = (4)(3)=12.A=(4)(3)=12.

Therefore,

ϕII=Ey A=(−5)(12)=−60 N m2/C.\phi_{II} = E_y\,A = (-5)(12) = -60\ \text{N m}^2/\text{C}.ϕII​=Ey​A=(−5)(12)=−60 N m2/C.
  1. Compute the required difference
ϕI−ϕII=0−(−60)=60.\phi_I - \phi_{II} = 0 - (-60)=60.ϕI​−ϕII​=0−(−60)=60.

So the required integer is

60.\boxed{60}.60​.
  1. Comparison with stored answer

Stored correct answer: −48-48−48

Our derived answer is 606060, which does not match. The stored answer appears inconsistent with the standard cuboid-face interpretation and the given field.

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