JEE Mains · Physics · STD 12 - 1. Electric charges and fields
A charge \('q'\) is placed at one corner of a cube as shown in figure. The flux of electrostatic field \(\overrightarrow{ E }\) through the shaded area is ...... .
- A \(\frac{ q }{4 \varepsilon_{0}}\)
- B \(\frac{ q }{24 \varepsilon_{0}}\)
- C \(\frac{ q }{48 \varepsilon_{0}}\)
- D \(\frac{ q }{8 \varepsilon_{0}}\)
Answer & Solution
Correct Answer
(B) \(\frac{ q }{24 \varepsilon_{0}}\)
Step-by-step Solution
Detailed explanation
flux through cube \(=\frac{ q }{8 \epsilon_{0}}\) flux through surfaces \(ABEH , ADGH , ABCD\) will be zero \(\phi( EFGH )=\phi( DCFG )=\phi( EBCF )=\frac{1}{3}\left(\frac{ q }{8 \epsilon_{0}}\right)\) \(=\frac{q}{24 \epsilon_{0}}\)
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