Electric Force versus Gravity and a Medium
For two particles, F_e/F_g = k|q_1q_2|/(Gm_1m_2). In a dielectric, k becomes k_0/K; immersed bodies have effective weight (rho_body-rho_fluid)Vg.
Why this shows up in the exam
Electric-to-gravity comparisons · Charged spheres in liquids · Equivalent separation in a dielectric
Learn the idea
Force ratios expose when gravity, permittivity, density, or buoyancy changes an electrostatic balance. Electric and gravitational forces can share the same inverse-square geometry, so their ratio may not depend on separation. Immersion also changes both electric force and apparent weight.
🧠 Memory hook: Same r squared cancels; a liquid changes both force and weight.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- F_e/F_g = k |q₁q₂|/(G m₁m₂) — distance-independent ratio for the same pair
- F_e,medium = F_e,vacuum/K — homogeneous linear dielectric
- W_eff = (rho_b-rho_f)Vg — weight minus buoyancy
How to approach it
- 1Write both competing forces
- 2Cancel common geometry
- 3Apply dielectric and buoyancy factors consistently
Common slip-ups that cost marks
- •Retaining r in the force ratio
- •Changing electric force but forgetting buoyancy
- •Using relative permittivity as an additive factor
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
Two point charges 1 microC and 2 microC are 1 m apart in vacuum. Take k = 9 x 10^9 SI. Find the force magnitude.
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