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Electrostatics question

2021 · 1 Sep · Shift 2 · Q47
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Electrostatics question

2021 · 1 Sep · Shift 2 · Q47

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A cube is placed inside an electric field, E→=150y2j^\overrightarrow E = 150{y^2}\widehat jE=150y2j​. The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is : JEE Main 2021 (Online) 1st September Evening Shift Physics - Electrostatics Question 132 English
  1. A
    3.8 ×\times× 10 −-− 11 C
  2. B
    8.3 ×\times× 10 −-− 11 C
  3. C
    3.8 ×\times× 10 −-− 12 C
  4. D
    8.3 ×\times× 10 −-− 12 C
View written solutionFree

Correct answer: B

  1. Use Gauss’s law

For any closed surface,

Φ=∮E⃗⋅dA⃗=Qencε0\Phi = \oint \vec E\cdot d\vec A = \frac{Q_{\text{enc}}}{\varepsilon_0}Φ=∮E⋅dA=ε0​Qenc​​

So,

Qenc=ε0ΦQ_{\text{enc}} = \varepsilon_0 \PhiQenc​=ε0​Φ
  1. Given electric field

The field is

E⃗=150y2 j^\vec E = 150y^2\,\hat jE=150y2j^​

This field is only along the yyy-direction and depends only on yyy.

Hence, flux through faces perpendicular to xxx and zzz is zero, because E⃗⊥dA⃗\vec E \perp d\vec AE⊥dA for those faces.

Only the two faces perpendicular to the yyy-axis contribute.

  1. Cube dimensions

Side of cube:

a=0.5 ma = 0.5\,\text{m}a=0.5m

Area of each face perpendicular to yyy-axis:

A=a2=(0.5)2=0.25 m2A = a^2 = (0.5)^2 = 0.25\,\text{m}^2A=a2=(0.5)2=0.25m2
  1. Flux through the two relevant faces

From the figure, the cube extends from y=0y=0y=0 to y=0.5 my=0.5\,\text{m}y=0.5m.

  • At the lower face, y=0y=0y=0:
E1=150(0)2=0E_1 = 150(0)^2 = 0E1​=150(0)2=0
  • At the upper face, y=0.5y=0.5y=0.5:
E2=150(0.5)2=150(0.25)=37.5 N/CE_2 = 150(0.5)^2 = 150(0.25)=37.5\,\text{N/C}E2​=150(0.5)2=150(0.25)=37.5N/C

Net outward flux:

Φ=E2A−E1A=(37.5)(0.25)−0=9.375\Phi = E_2A - E_1A = (37.5)(0.25)-0 = 9.375Φ=E2​A−E1​A=(37.5)(0.25)−0=9.375

So,

Φ=9.375 N m2/C\Phi = 9.375\,\text{N m}^2/\text{C}Φ=9.375N m2/C
  1. Charge enclosed

Using

Qenc=ε0ΦQ_{\text{enc}} = \varepsilon_0 \PhiQenc​=ε0​Φ

with

ε0=8.85×10−12 C2/N m2\varepsilon_0 = 8.85\times 10^{-12}\,\text{C}^2/\text{N m}^2ε0​=8.85×10−12C2/N m2

we get

Qenc=8.85×10−12×9.375Q_{\text{enc}} = 8.85\times 10^{-12}\times 9.375Qenc​=8.85×10−12×9.375 Qenc≈8.30×10−11 CQ_{\text{enc}} \approx 8.30\times 10^{-11}\,\text{C}Qenc​≈8.30×10−11C
  1. Match with options
Qenc=8.3×10−11 CQ_{\text{enc}} = 8.3\times 10^{-11}\,\text{C}Qenc​=8.3×10−11C

So the correct option is:

B

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