JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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 \(.....\times 10^{-11} {C}\)

- A \(3.8\)
- B \(8.3\)
- C \(0.38\)
- D \(830\)
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…
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