Question 703: Electrostats & Capacitance

Question

The electric field in a region is given by $\overrightarrow{\mathrm{E}}=(2 \hat{i}+4 \hat{j}+6 \hat{k}) 	imes 10^3 \mathrm{~N} / \mathrm{C}$. The flux of the field through a rectangular surface parallel to $x-z$ plane is $6.0 \,\mathrm{Nm}^2 \mathrm{C}^{-1}$. The area of the surface is _____________ $\mathrm{cm}^2$.

Accepted Answer

To calculate the area of the surface through which the electric flux is determined, we use the given electric field and the specified surface orientation. The electric field is expressed as $\overrightarrow{\mathrm{E}} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) 	imes 10^3 \, \mathrm{N/C}$.
Since the surface is parallel to the $x-z$ plane, its area vector $\overrightarrow{\mathrm{A}}$ is directed along the $y$-axis, making it $\mathrm{~A} \hat{\mathrm{j}}$.
The electric flux $\phi$ through the surface is given by:
$ \phi = \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{~A}} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) 	imes 10^3 \cdot \mathrm{~A} \hat{\mathrm{j}} $
Calculating the dot product, the only component contributing to the flux is the $y$-component:
$ \phi = (4 	imes 10^3) \mathrm{~A} $
Given that the electric flux $\phi$ is $6.0 \, \mathrm{Nm}^2/\mathrm{C}$, we can equate and solve for $\mathrm{A}$:
$ 6 = 4 	imes 10^3 \mathrm{~A} $
$ \mathrm{A} = \frac{6}{4 	imes 10^3} = 1.5 	imes 10^{-3} \, \mathrm{m}^2 $
Converting this area from square meters to square centimeters:
$ \mathrm{A} = 1.5 	imes 10^{-3} \, \mathrm{m}^2 = 15 \, \mathrm{cm}^2 $

Electrostats & Capacitance

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