The electric field \(\vec{E}_{1}\) at one face of a parallelopiped is uniform over the entire face and is…

The electric field \(\vec{E}_{1}\) at one face of a parallelopiped is uniform over the entire face and is directed out of the face. At opposite face, the electric field \(\vec{E}_{2}\) is also uniform over the entire face and is directed into that face (as shown in figure). The two faces in question are inclined at \(30^{\circ}\) from the horizontal, while \(\overrightarrow{E_{1}}\) and \(\overrightarrow{E_{2}}\) (both horizontal) have magnitudes of \(2.50 \times 10^{4} \mathrm{~N} / \mathrm{C}\) and \(7.00 \times 10^{4}\) N.C. Assuming that no other electric field lines cross the surfaces of the parallelopiped, determine the net charge contained within.
  1. \(-67.5 \varepsilon_{0} \mathrm{C}\)
  2. \(37.5 \varepsilon_{0} \mathrm{C}\)
  3. \(105 \varepsilon_{0} \mathrm{C}\)
  4. \(-105 \varepsilon_{0} \mathrm{C}\)

Solution

To find the charge enclosed, we need the flux through the parallolepiped:
\(\begin{array}{l}
\varphi_{1}=\mathrm{AE}_{1} \cos 60^{\circ} \\
=(0.0500 \mathrm{~m})(0.0600 \mathrm{~m})\left(2.50 \times 10^{4} \mathrm{NC}^{-1} \cos 60^{\circ}\right. \\
=37.5 \mathrm{Nm}^{2} \mathrm{C}^{-1} \\
\varphi_{2}=\mathrm{AE}_{2} \cos 120^{\circ} \\
=(0.0500 \mathrm{~m})(0.0600 \mathrm{~m})\left(7.00 \times 10^{4} \mathrm{NC}^{-1} \cos 60^{\circ}\right. \\
=-105 \mathrm{Nm}^{2} \mathrm{C}^{-1}
\end{array}\)
So, the total flux is
\(\begin{array}{l}
\varphi=\varphi_{1}+\varphi_{2}=(37.5-105) \mathrm{Nm}^{2} \mathrm{C}^{-1}=-67.5 \mathrm{Nm}^{2} \mathrm{C}^{-1} \\
q=\varphi \varepsilon_{0}=\left(-67.5 \mathrm{Nm}^{2} / \mathrm{C} \varepsilon_{0}\right)=-5.97 \times 10^{-10} \mathrm{C}
\end{array}\)
There must be a net charge (negative) in the parallelopiped since there is a net flux flowing into the surface. Also, there must be an external field or all lines would point toward the slab. .

Asked in: JEE Mains - Electrostatics - Chapter Test

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