The total flux associated with given cube will be-
where $^{'} a^{'}$ is side of cube : $(\frac{1}{ϵ_{0}} = 4 π \times 9 \times 10^{9} S I u n i t)$

$162 π \times 10^{- 3} N m^{2} / C$
$162 π \times 10^{3} N m^{2} / C$
$162 π \times 10^{- 6} N m^{2} / C$
$162 π \times 10^{6} N m^{2} / C$

Question

The total flux associated with given cube will be-
where $^{'} a^{'}$ is side of cube : $(\frac{1}{ϵ_{0}} = 4 π \times 9 \times 10^{9} S I u n i t)$

$162 π \times 10^{- 3} N m^{2} / C$
$162 π \times 10^{3} N m^{2} / C$
$162 π \times 10^{- 6} N m^{2} / C$
$162 π \times 10^{6} N m^{2} / C$

Toppr Admin · Accepted Answer

Total flux associated with given cube-xA0-1-x3F5-0-4-x3C0-xD7-9-xD7-109SI-q1-q2-q3-q4-x3C0-xD7-103Nm2-C-4-xD7-1-xD7-4-xD7-3-xD7-x3C0-xD7-103-xD7-10-x2212-6Nm2-C-162-xD7-x3C0-xD7-103Nm2-C

The total flux associated with given cube will be- where ′a′ is side of cube : (1ϵ0=4π×9×109 SI unit)162π×10−3Nm2/C162π×103Nm2/C162π×10−6Nm2/C162π×106Nm2/C

Total flux associated with given cube 1ϵ0=4π×9×109SI(q1+q2+q3+q4)π×103Nm2/C=(4×1×4×3)×π×103×10−6Nm2/C=162×π×103Nm2/C

The total flux associated with given cube will be-
where $^{'} a^{'}$ is side of cube : $(\frac{1}{ϵ_{0}} = 4 π \times 9 \times 10^{9} S I u n i t)$

$162 π \times 10^{- 3} N m^{2} / C$
$162 π \times 10^{3} N m^{2} / C$
$162 π \times 10^{- 6} N m^{2} / C$
$162 π \times 10^{6} N m^{2} / C$

Total flux associated with given cube $\frac{1}{ϵ_{0}} = 4 π \times 9 \times 10^{9} S I$
$(q_{1} + q_{2} + q_{3} + q_{4}) π \times 10^{3} N m^{2} / C$
$= (4 \times 1 \times 4 \times 3) \times π \times 10^{3} \times 10^{- 6} N m^{2} / C$
$= 162 \times π \times 10^{3} N m^{2} / C$