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Electrostatics question

2023 · 30 Jan · Shift 1 · Q49
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  5. /2023 · 30 Jan · Shift 1 · Q49

Electrostatics question

2023 · 30 Jan · Shift 1 · Q49

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Electric field in a certain region is given by E→=(Ax2i^+By3j^). The SI unit of A and B\overrightarrow{\mathrm{E}}=\left(\frac{\mathrm{A}}{x^{2}} \hat{i}+\frac{\mathrm{B}}{y^{3}} \hat{j}\right) \text {. The } \mathrm{SI} \text { unit of } \mathrm{A} \text { and } \mathrm{B}E=(x2A​i^+y3B​j^​). The SI unit of A and B are :
  1. A
    Nm2C;Nm3C\mathrm{Nm}^{2} \mathrm{C} ; \mathrm{Nm}^{3} \mathrm{C}Nm2C;Nm3C
  2. B
    Nm3C−1;Nm2C−1\mathrm{Nm}^{3} \mathrm{C}^{-1} ; \mathrm{Nm}^{2} \mathrm{C}^{-1}Nm3C−1;Nm2C−1
  3. C
    Nm3C;Nm2C\mathrm{Nm}^{3} \mathrm{C} ; \mathrm{Nm}^{2} \mathrm{C}Nm3C;Nm2C
  4. D
    Nm2C−1;Nm3C−1\mathrm{Nm}^{2} \mathrm{C}^{-1} ; \mathrm{Nm}^{3} \mathrm{C}^{-1}Nm2C−1;Nm3C−1
View written solutionFree

Correct answer: D

  1. The electric field has SI unit [E]=N C−1.[E] = \mathrm{N\,C^{-1}}.[E]=NC−1.

  2. Given E⃗=(Ax2i^+By3j^).\vec E = \left(\frac{A}{x^2}\hat i + \frac{B}{y^3}\hat j\right).E=(x2A​i^+y3B​j^​). Since each component of electric field must have unit N C−1\mathrm{N\,C^{-1}}NC−1, we compare units term-by-term.

  3. For the xxx-component: Ax2∼N C−1.\frac{A}{x^2} \sim \mathrm{N\,C^{-1}}.x2A​∼NC−1. Since [x]=m[x] = \mathrm{m}[x]=m, [A]=[E][x2]=(N C−1)(m2)=N m2 C−1.[A] = [E][x^2] = \left(\mathrm{N\,C^{-1}}\right)(\mathrm{m^2}) = \mathrm{N\,m^2\,C^{-1}}.[A]=[E][x2]=(NC−1)(m2)=Nm2C−1.

  4. For the yyy-component: By3∼N C−1.\frac{B}{y^3} \sim \mathrm{N\,C^{-1}}.y3B​∼NC−1. Since [y]=m[y] = \mathrm{m}[y]=m, [B]=[E][y3]=(N C−1)(m3)=N m3 C−1.[B] = [E][y^3] = \left(\mathrm{N\,C^{-1}}\right)(\mathrm{m^3}) = \mathrm{N\,m^3\,C^{-1}}.[B]=[E][y3]=(NC−1)(m3)=Nm3C−1.

  5. Therefore, A:N m2 C−1,B:N m3 C−1.A: \mathrm{N\,m^2\,C^{-1}}, \qquad B: \mathrm{N\,m^3\,C^{-1}}.A:Nm2C−1,B:Nm3C−1.

  6. Checking options:

    • A: incorrect
    • B: incorrect
    • C: incorrect
    • D: correct

Hence, the correct option is D.

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