mMCQ.

Navigation Menu

Step into mMCQ.

Launch mMCQ. diagnostic

Explore mMCQ.

MDCAT prepFree DiagnosticPricing & SubscribeSign in

Resources

Terms & Conditions

mMCQ.

© 2021 - 2025 mMCQ.All rights reserved.

WhatsApp
A Levels Physics (9702)•9702/11/O/N/21
Question 31 from 9702/11/O/N/21

Explanation

Ratio of electric field magnitudes determines force ratio on electron

Steps:

  • Potential difference upper region (X): 500 V - 200 V = 300 V, distance 0.02 m, so E_X = 300 / 0.02 = 15,000 V/m.
  • Potential difference lower region (Y): 200 V - 0 V = 200 V, distance 0.01 m, so E_Y = 200 / 0.01 = 20,000 V/m.
  • Force magnitude F = eE, so ratio F_X / F_Y = E_X / E_Y = 15,000 / 20,000 = 0.75.

Why A is correct:

  • Matches the ratio F_X / F_Y = 0.75 from E = ΔV/d formula for uniform fields between charged plates.

Why the others are wrong:

  • B 1.00: Assumes equal fields, ignoring different ΔV and d ratios.
  • C 1.25: Swaps regions, yielding E_Y / E_X = 20,000 / 15,000 ≈ 1.33 (not exact).
  • D 1.50: Ignores distances, using raw ΔV ratio 300/200 = 1.5.

Final answer: A

Topic: Uniform electric fields

Practice more A Levels Physics (9702) questions on mMCQ.me