JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 9)
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 :
_25th_February_Evening_Shift_en_9_1.png)
$${q \over {24{\varepsilon _0}}}$$
$${q \over {48{\varepsilon _0}}}$$
$${q \over {4{\varepsilon _0}}}$$
$${q \over {8{\varepsilon _0}}}$$
Explanation
_25th_February_Evening_Shift_en_9_2.png)
Flux through cube, $$\phi = {q \over {8{\varepsilon _0}}}$$
Flux through surfaces ABEH, ADGH, ABCD will be zero.
$$\phi $$(EFGH) = $$\phi $$(DCFG) = $$\phi $$(EBCF) = $${1 \over 3}\left( {{q \over {8{\varepsilon _0}}}} \right)$$
= $${q \over {24{\varepsilon _0}}}$$
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