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1. Electric Charges and Fields
normal

$q$ વિદ્યુતતારને એક બંધ ઘનના કેન્દ્ર આગળ મૂકવામાં આવે છે ઘનના કોઈ પણ એક છેડામાંથી બહાર આવતું ફલક્સ ....... હશે.

A

$Q/6$ $\varepsilon_0$

B

$Q/3$ $\varepsilon_0$

C

$Q/$$\varepsilon_0$

D

$Q/4$ $\varepsilon_0$

Solution

Total flux coming out $=\frac{\text { Qenclosed }}{\epsilon_0}$

$\int E . ds =\frac{ Q \times 10^{-6}}{\epsilon_0}$ (guass law)

As $Q$ is placed at center, flux coming out of all sides will be equal.

$\therefore \text { flux from one of the side }=\frac{ Q \times 10^{-6}}{6 \epsilon_0}$

$=\frac{ Q \times 10^{-6}}{6 \epsilon_0}$

Standard 12
Physics

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