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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2021
  • Hard
  • MCQ

A cube is placed inside an electric field, \(\overrightarrow{{E}}=150\, {y}^{2}\, \hat{{j}}\). 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 \(.....\times 10^{-11} {C}\)

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Answer & Solution

Correct Answer

(B) \(8.3\)

Step-by-step Solution

Detailed explanation

As electric field is in \(y-\)direction so electric flux is only due to top and bottom surface

Bottom surface \({y}=0\)

\(\Rightarrow {E}=0 \Rightarrow \phi=0\)

Top surface \({y}=0.5\, {m}\)

\(\Rightarrow {E}=150(0.5)^{2}=\frac{150}{4}\)

Now flux \(\phi={EA}=\frac{150}{4}(.5)^{2}=\frac{150}{16}\)

By Gauss's law \(\phi=\frac{Q_{\text {in }}}{\epsilon_{0}}\)

\(\frac{150}{16}=\frac{{Q}_{\text {in }}}{\epsilon_{0}}\)

\({Q}_{\text {in }}=\frac{150}{16} \times 8.85 \times 10^{-12}=8.3 \times 10^{-11} \,{C}\)

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