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Six point charges, each of the same magnitude q , are arranged in different manners as shown in Column-II. In each case, a point M and a line P Q passing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line

Options

  1. A(A) q,t, (B) q,s, (C) p,q,t, (D) r,s,t
  2. B(A) q,r,s, (B) r,s, (C) p,r, (D) r,s,t
  3. C(A) p,r,s, (B) r,s, (C) p,q,t, (D) r,s
  4. D(A) p,r, (B) q,s, (C) p,r,t, (D) r,s

Correct answer

C. (A) p,r,s, (B) r,s, (C) p,q,t, (D) r,s

Step-by-step solution

Concise justification (one line per match): - A: (E=0 p, r, s ). Each of these arrangements is point-symmetric about (M ) so the vector sum of electric fields cancels at (M ) : (p) alternating charges on a regular hexagon, (r) concentric rings with symmetric charge placement, (s) symmetrical rectangle with charges at corners/midpoints. - B: (V 0 r ), s. Potential is a scalar sum and need not cancel even if fields cancel; in ( (r )) and ( (s )) the positive and negative charges sit at different radii/positions so th

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