Electric field halves when plate separation doubles at constant voltage

Steps:

• Electric field E between parallel plates is E = V/d, where V is battery voltage and d is separation.
• Original force F = qE = qV/d, with q as particle charge.
• Doubling separation to 2d gives new field E' = V/(2d) = E/2.
• New force F' = qE' = q(V/(2d)) = F/2.

Why B is correct:

• Force F = qE and E = V/d, so doubling d halves E and thus halves F at constant V.

Why the others are wrong:

• A: Doubling d halves E once, not twice to quarter F.
• C: E changes inversely with d, so F changes.
• D: Doubling d decreases E, reducing F.

Final answer: B