11Ans 1) C, D
Under electrostatic conditions , all points lying on the conductor are in the same potential. Therefore, potential at A= potential at B.
Frm Gauss's theorm, total flux through the surface of the cavity will be q/ε0.
1) (Multiple Answers Correct)
An elliptical cavity is carved within a perfect conductor. A +ve charge "q" is placed at the centre of the cavity. Then-
A) Elec. field near A = Elec. field near B (in the cavity).
B) Charge density at A = Charge Density at B.
C) Potential at A = Potential at B
D) Total Elec. Field Flux through the surface of the cavity is q/ε0.
2)Assertion-Reason Type
STATEMENT I- On going away frm a point charge or a small elec. dipole, elec. field decrease at the same rate for both the cases.
STATEMENT II- Electric field is inversely proportional to square of distance for a point charge.
A) Both correct, II explains I.
B) Both correct, II dosen't explain I.
C) I correct, II wrong.
D) I wrong, II correct.
E) Both wrong.
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7 Answers
11For the 1st one, its A,C,D for sure....
Thinking abt B...not too sure.
D) for the 2nd qsn.
1D certainly for 2(field inversely proportional to r3 for dipole and r2 for point charge).
A,C,D for 1?
not confident about B
11
13Fr#1 - [In Qn1 - A & B are ON the Cavity surface.] [1]
So, the answers given are- 1)C,D ...2)C
.... Wht i start with is- In the +nce of an external elec. field, charges arrange themselves in such a way tht net field inside the conductor is Zero.
Then, how can there be any net flux coming out thro' the cavity ?
1yeah but ur talking about a Solid conductor..in the given case..where it is hollow at a certain part(the cavity), there is a net flux
D is a right answer..but why C??
13Yep, it will be D) only, answer given is wrong, so v shouldn't confuse ourselves over this... [1]