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Question: The electric field in a region is given by \[\vec E = \dfrac{A}{{{x^3}}}\hat i + By\hat j + C{z^2}\h...

The electric field in a region is given by E⃗=Ax3i^+Byj^+Cz2k^\vec E = \dfrac{A}{{{x^3}}}\hat i + By\hat j + C{z^2}\hat k. The S.I units of A, B and C respectively.
A. Nm3C,Vm−2,N(m2C)−1\dfrac{{N{m^3}}}{C},V{m^{ - 2}},N{({m^2}C)^{ - 1}}
B. Vm2,Vm−1,N(m2C)−1V{m^2},V{m^{ - 1}},N{({m^2}C)^{ - 1}}
C. Vm−2,Vm−1,NCm−2V{m^{ - 2}},V{m^{ - 1}},NC{m^{ - 2}}
D. Vm−1,Nm3C,NCm−1V{m^{ - 1}},\dfrac{{N{m^3}}}{C},NC{m^{ - 1}}

Explanation

Solution

We need to know the Possible SI units of Electric Field. Electric Field of a region can be expressed in form as either V/m2 or as N/C. Using Dimensional Analysis, we can solve for SI Unit.

Complete Step By Step Solution:
SI unit of Strength of an Electric field can be represented in many formats. Such as Vm−1V{m^{ - 1}} and NC−1N{C^{ - 1}}, where V is Voltage per meter or Newton per capacitance.
For the given equation we can use dimensional analysis to solve the equation.
E⃗=Ax3i^+Byj^+Cz2k^\vec E = \dfrac{A}{{{x^3}}}\hat i + By\hat j + C{z^2}\hat k
We know that Force of a charge=qE = qE
Hence, E=Fq=Kg×ms2×C=NCE = \dfrac{F}{q} = \dfrac{{Kg \times m}}{{{s^2} \times C}} = \dfrac{N}{C}
So the possible SI unit of A can be,
Ax3i^\dfrac{A}{{{x^3}}}\hat i= NC\dfrac{N}{C}
⇒Am3=NC\Rightarrow \dfrac{A}{{{m^3}}} = \dfrac{N}{C}

Note: x is the distance of the point, so the SI unit of distance is considered as m.
⇒A=Nm3C\Rightarrow A = \dfrac{{N{m^3}}}{C}
Similarly, for B, we consider
Byj^By\hat j= NC\dfrac{N}{C}= V/mV/m
⇒B×m=NC=Vm\Rightarrow B \times {m^{}} = \dfrac{N}{C} = \dfrac{V}{m}
⇒B=NmC=Vm2\Rightarrow B = \dfrac{N}{{mC}} = \dfrac{V}{{{m^2}}}
Similarly for equation C,
Cz2k^C{z^2}\hat k == NC=Vm\dfrac{N}{C} = \dfrac{V}{m}
⇒C×m2=NC=Vm\Rightarrow C \times {m^2} = \dfrac{N}{C} = \dfrac{V}{m}
⇒C=Nm2C=Vm3\Rightarrow C = \dfrac{N}{{{m^2}C}} = \dfrac{V}{{{m^3}}}
Therefore, SI unit of A=Nm3CA = \dfrac{{N{m^3}}}{C}, B=NmC=Vm2B = \dfrac{N}{{mC}} = \dfrac{V}{{{m^2}}} and C=Nm2C=Vm3C = \dfrac{N}{{{m^2}C}} = \dfrac{V}{{{m^3}}}.

According to the options, Option A is the right answer.

Note:
Electric field strength is defined as the intensity of the electric field in a particular region. The strength of the electric field generated by a source charge Q is inversely proportional to the square of the distance from the source. It is also known as inverse square law. The SI unit of Electric field is given as Newton per Coulomb or Volt per meter.