Question

A very small object with mass 8.10×10−9 kg and positive charge 6.20×10−9 C is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density 5.90×10−8C/m2. The object is initially 0.590 m from the sheet.

What initial speed must the object have in order for its closest distance of approach to the sheet to be 0.330 m ?

An electron is released from rest at a distance of 0.530 m from a large insulating sheet of charge that has uniform surface charge density 3.30×10⁻¹²C/m2 .

What is the speed of the electron when it is 2.00×10⁻² m from the sheet?

Answer 1

The field from an (effectively infinite) sheet of charge is: E = σ/(2ε₀)

The force on the particle is F=qE acting away from the sheet: F = qE = qσ/(2ε₀)

If v is the initial velocity, the initial kinetic energy = ½mv².

At the point of closest approach, the kinetic energy is instantaneously zero (the particle has come to rest and is about to change direction, a bit like at ball thrown upwards when it reaches max. height).

The *change* in kinetic energy is -½mv² (negative as KE has been reduced).

The particle moves a distance d = 0.590 - 0.330 = 0.260m. The direction is against the force. Therefore the work done by the force is negative. So we treat either F as positive and d as negative, or vice versa.
Fd = Δ(KE) (work-energy theorem)
-(qσ/(2ε₀))d = -½mv²
qσd/ε₀ = mv²
v² = qσd/(ε₀m)
v = √[qσd/(ε₀m)] = √(6.2*10^-9*5.9*10^-8*0.26)/(8.85*10^-12*8.1*10^-9) = 36.4 m/s

2)

field E = σ / 2ε₀ = 3.3E-12C/m² / 2*8.85e-12 C²/N·m² = 0.1864 N/C

work = q*E*d = 1.6E-19C * 0.1864N/C * (0.530m - 0.02m) = 1.52E-20 J

work becomes KE
1.52E-20 J = ½ * 9.11E-31kg * v²
v = 1.827E5 m/s