For two equal positive charges Q fixed at x = ±a, a positively charged particle moves on the perpendicular bisector y-axis. Its two gravitational-electrostatic equilibrium points are OA = a/(2√3) and OB = 3a/2. Which statements are correct for 0 < y < OB?
Select all that apply

Question

For two equal positive charges Q fixed at x = ±a, a positively charged particle moves on the perpendicular bisector y-axis. Its two gravitational-electrostatic equilibrium points are OA = a/(2√3) and OB = 3a/2. Which statements are correct for 0 < y < OB?

Select all that apply

G1 Physics · Accepted Answer

The electric field is E(y)=2kQy/(a²+y²)³ᐟ². Differentiate: dE/dy = 0 ⇒ a² - 2y² = 0 ⇒ y = a/√2. Since a/(2√3) < a/√2 < 3a/2, the maximum lies between A and B. The electric force qE is maximum there. For fixed gravity, the upward excess qE-mg is also maximized there. The field is not monotonic over O to B.

For two equal positive charges Q fixed at x = ±a, a positively charged particle moves on the perpendicular bisector y-axis. Its two gravitational-electrostatic equilibrium points are OA = a/(2√3) and OB = 3a/2. Which statements are correct for 0 &lt; y &lt; OB?

For two equal positive charges Q fixed at x = ±a, a positively charged particle moves on the perpendicular bisector y-axis. Its two gravitational-electrostatic equilibrium points are OA = a/(2√3) and OB = 3a/2. Which statements are correct for 0 < y < OB?