a point charge q is placed at the centre of one of the faces of an imaginary hollow cube. the electric flux through the opposite face is
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ACCORDING TO THE GIVEN PROBLEM, THE CAHRGE IS PLACED AT THE CORNER OF THE CUBE AS SHOWN IN FIGHURE_1,
NOW CONSIDER THE ANOTHER FIGURE_2 IN WHICH THE WHOLE CHARGE IS ENCLOSED IN THE FOUR SUCH CUBE,
NOW CALCULATING THE FLUX THROUGH THE ALL CUBES AS TOTAL:
ϕ
e
=
ϵ
0
q
=EA
WHERE A=6∗(2a)
2
FOR ONLY ONE SIDE OF FIRST CUBE,
AREA WILL BE A
0
=a
2
FROM ABOVE DATA FLUX THROW THE SINGLE FACE (AREA = A
0
) IS GIVEN BY:
ϕ
e
=
ϵ
0
Q
=EA
0
WHERE Q=
24
q
SO THE FLUX THROUGH THROUGH THE SINGLE FACE IS
24ϵ
0
q
;
HENCDE THE OPTION (D) IS CORRECT.
Explanation:
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