Question

A small sphere with mass 5.00.×10−7kg and charge +3.00μC is released from rest a distance of 0.600 m above a large horizontal insulating sheet of charge that has uniform surface charge density σ=+8.00pC/m2. Using energy methods, calculate the speed of the sphere when it is 0.100 m above the sheet of charge?

Answer 1

Answer:

v = 5201m/s.

Explanation:

The Work done on charge is W = E×q×d = σ/(2ε₀) x q x d = {(8.00 x 10⁻¹²)/(2 x 8.854187 x 10⁻¹²)} x 3.00 x 10⁻⁶ x (0.600 - 0.100) = 6.765J.

KE of sphere = 0.5mv² = 0.5 x 5.00 x 10⁻⁷v² = work done by E-field on charge during its fall = 6.765

2.5×10^-7V² = 6.765

V = √6.765/2.5×10^-7

v = 5201m/s.

Answer 2

Given:

m = 5.00x10^-7 kg
q = 3.00μC

To determine the velocity, use this formula

v = √(2qΔx/m)

Now, solve for the velocity, substitute the given values to the equation

v = √(2(3.00μC * 0.600m/5.00x10^-7 kg)

Solve for V and this is the velocity of your sphere in condition 1.