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1. Electric Charges and Fields
medium
The electric flux passing through the cube for the given arrangement of charges placed at the corners of the cube (as shown in the figure) is
A
$\phi = \frac{1}{{2{ \in _0}}}$
B
$\phi = \frac{{ - 1}}{{2{ \in _0}}}$
C
$\phi = \frac{{ - 1}}{{{ \in _0}}}$
D
$\phi = \frac{1}{{{ \in _0}}}$
Solution
$\phi = \frac{{{{\rm{q}}_{{\rm{net}}}}}}{{{ \in _0}}}$
for a charge placed at corner of cube
$\phi=\frac{q_{\text {net }}}{8 \in_{0}}$
$\therefore $ thus for given system
$\phi = \frac{{(1 – 2 + 3 – 4 – 6 + 7 + 5 – 8)}}{{8{ \in _0}}}$
$\phi = \frac{{ – 1}}{{2{ \in _0}}}$
Standard 12
Physics
