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Asked in Re-NEET 2026 · Conductors and electrostatic shielding
A lies in the cavity, close to Q: the field of Q (and of the charge induced on the cavity wall) is not zero there, so E_A eq0.
The metal shields the outside from the cavity: charge -Q collects on the cavity wall and +Q spreads uniformly over the outer surface.
Outside, the field is that of +Q at the centre: E=(kQ)/(r²), the same at B and C, which are equally far from the centre.
So E_A eq0 and E_B=E_C.
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