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

The speed of the sphere when it is 0.100 m above the sheet of charge is approximately 1.98 m/s.

Step-by-step explanation:

To find the speed of the sphere when it is 0.100 m above the sheet of charge, we can use energy conservation. The initial energy of the sphere is entirely gravitational potential energy, which is given by:

PEi = mgh = (5.00×10-7 kg)(9.8 m/s2)(0.600 m) = 2.94×10-7 J

The final energy of the sphere is the sum of gravitational potential energy and electric potential energy:

PEf = mgh + qV

where q is the charge on the sphere and V is the electric potential of the sheet of charge.

Using the given charge on the sphere (+3.00μC) and electric potential of the sheet of charge (+8.00pC/m2), we can calculate the final energy:

PEf = (5.00×10-7 kg)(9.8 m/s2)(0.100 m) + (+3.00×10-6 C)(8.00×10-6 C/m2) = 1.96×10-7 J

Since energy is conserved, we can equate the initial and final energies and solve for the speed of the sphere:

PEi = PEf

2.94×10-7 J = 1.96×10-7 J + (1/2)mv2

Simplifying the equation, we find:

v2 = (2(2.94×10-7 J - 1.96×10-7 J))/m = (1.96×10-7 J)/m

Plugging in the given mass of the sphere (5.00×10-7 kg), we can calculate the speed:

v2 = (1.96×10-7 J)/(5.00×10-7 kg) = 3.92 m2/s2

v = √(3.92 m2/s2) ≈ 1.98 m/s

Answer 2

v = 5201m/s.

Step-by-step 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.