1Gauss' Law is not required...simple logic is enough !
A point charge q is placed on one vertex of a cube. What part of the charge is inside the cube?
Answer in 1/8th. Why??
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11 Answers
1yes I did use gauss law to find the field, and the answr is q/8ε . But why is it that 1/8th of the charge will be inside the cube if the charge is considered spherical?? this part is NOT gauss's law
1its much as we did in solid state,madhumitha.
altogether 8 cubes will be needed to completely cover the charge at their common vertex.
thus each cube encloses 1/8 of it[1]
1yes , really crazy to use gauss law ...
she hasnt asked flux ....
@arka yup thats the reason
@madhimitha even if it fux askd
the answer will be q/6ε
1This question is based on symmetry going by it the answer should be " 1/8q "
1thanks arka
Debashree, I didnt do Gauss's law to find the amt of charge enclosed..i used gauss to find the flux..which is the second part of the question
gr8freams, I dont see how the answer is q/6ε ..i got q/8ε
1construct 7 more cubes to form an imaginary cube such that the charge is noe in body centre...now u can calculate flux thru entire big cube =q/ε
symmetry says each face must have equ al amount of field lines passing thru them
flux to each face =q/6ε
hence flux trhu one side of small cube is
q/24ε
EDIT : THIS FFLUX IS THE FLUX PASSING THRU NON ADJACENT CORNER TO THE VERTEX WER THE CHARGE IS KEPT
1well, wat about this>?
consider the cube with charge Q placed @ its vertex. 1/8th of the charge is insude the cube. so flux through cube is jsut Q/8ε
1sorry guys , i thought it was another question
i thought question was
if a charge is kept at coner find flux passing thru the non adjacent face