help in physics

Question

In a Millikan-type experiment, two horizontal charged plates are 2.5 cm apart. A latex sphere of 1.3 x 10^-15 kg remains stationary between the plates when the potential difference between the plates is 400 V, with the upper plate charged positive.

a) What is the type of charge on the sphere?

(give your answer  either positive, negative or neutral)

b) The electric field intensity between the plates is ___ x10⁴ V/m

(Give your answer  in correct significant digits with no units)

c) What is the direction of the electric field?

(Give your answer i either up, down, north, south, east or west)

d) The charge on the sphere is ___ x 10^-19 C

Answer 1

a)

Upper plate is positively charged.

The top plate would repel the sphere , so that the sphere must be charged oppositely as the upper plate charged.

Sphere is charged Negatively.

Answer 2

b)

Distance between plates d = 2.5 cm = 2.5 × 10^-2 = 0.025 m.

Potential difference between the plates v = 400 V.

The electric field intensity E = v/d

Substitute : d = 0.025 , v = 400.

E = 400 / 0.025 = 16000 = 1.6 × 10⁴

The electric field intensity between the plates is 1.6 × 10⁴ V/m.

Answer 3

c)

A negative charge of the sphere will be attracted to the upper positive plate with an upward force.

And it is opposite to the force of gravity.

So that the direction of the electric field is Upwards.

Answer 4

d)

Latex sphere mass m = 1.3 × 10^-15 kg.

The electric field intensity E = 1.6 × 10⁴ V/m.

The force of gravity acting upon the sphere F_gravity = mg

The electric force on the sphere F_electric = qE

The sphere is stationary, then the force of gravity is equals the electric force on the sphere.

F_gravity = F_electric

mg = qE

q = mg/E

q = ( 1.3 × 10^-15 )(9.8) / (1.6 × 10⁴ )

q = (1.3× 9.8)(0.625)10^-15-4

q = 7.9625 ×10^-19 C

The charge on the sphere is 7.9625 x 10^-19 C.