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A spherical metal shell A of radius RA a...

A spherical metal shell A of radius `R_A` and a solid metal sphere B of radius `R_B(ltR_A)` are kept far apart and each is given charge `'+Q'`. Now they are connected by a thin metal wire. Then

A

`E_(A)^(" in side") =0`

B

`Q_(A) gt Q_(B)`

C

`(sigma_(A))/(simga_(B)) = (R_(B))/(R_(A))`

D

`E_(A)^("on surface") lt E_(B)^("on surface ")`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
Inside a conducting shell electric field is always zero. Therefore, option (A) is correct. When the two are connected, their potential is become the same.
`therefore V_(A)=V_(B) or Q_(A)/R_(A)=Q_(B)/(R_(B) (V=1/(4piepsi_(0) Q/R)`
Since, `R_(A) gt R_(B) therefore Q_(A) gt Q_(B)`
Potential is also equal to, `V=(sigmaR)/(epsi_(0)) , V_(A)=V_(B) therefore sigma_(A) R_(A) =sigma_(B) R_(B) or (sigma_(A))/(sigma_(B))=(R_(B))/(R_(A))` or `sigma_(A) lt sigma_(B)`
Electric field on surface, `E=(sigma)/(epsi_(0)), or E alpha sigma ," Since "sigma_(A) lt sigma_(B) therefore E_(A) lt E_(B)`
Hence, options (A), (B), (C) and (D) are correct.
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