Physics question help please?

Question

Question 1 

If it takes 4.87 J of energy to move a 2.12 C charge in an electric field, then the potential difference is _.__ V. 


Question 2 

The work required to move a 6.42 mC charge across a potential difference of 222 V is _.__ J. 


Question 3 

Two charged plates are separated by a distance of 6.00 mm. If the potential difference between the plates is (5.80x10^3) V, the acceleration of a proton placed in the field would be _.__ x10^ m/s2. 

Question 4 

Two charged plates are 5.00 cm apart and have a electric field strength of 606 V/m. If the plates are moved 7.00 cm apart, and the potential difference held constant, the electric field strength becomes ___ V/m. 

Question 5 

A small foam sphere has a mass of 2.70x10-3 kg and is suspended from a light thread between two charged plates that are 12.5 cm apart. When a potential difference of (5.05x10^2) V is applied to the plates, the ball moves the right creating an angle of 0.613 degrees E of S (shown below). The charge on the foam sphere is _.__ x 10^ C.

Answer 1

(1)

The Energy = 4.87 J

Charge = 2.12 C

Potential Difference = Work / charge

Potential Difference V = 4.87 / 2.12

V = 2.297 volts

Therefore the Potential Difference is 2.297 v.

Answer 2

(2)

Charge of the Particle is 6.42 mC = 6.42 * 10^-3 C.

Potential Difference between the plates = 222 v

Work = charge * Potential Difference

W = 6.42 * 10^-3 * 222

W = 1.425 J

Therefore Work done in moving a charged particle is 1.425 J.

Answer 3

(3)

The distance between two charged plates = 6.0 mm 0.006 m.

Potential Difference between to the planes = 5.80 * 10³.

Electric Potential E = Potential Difference / distance

E = (5.80 * 10³) / 0.006

E = 966.67 kJ.

Now accerleration = qE/m

We know that q = 1.602 * 10^-19 C

Mass of the Proton = 1.6762 * 10^-27 kg.

Accerleration = (1.602*10^-19 * 96666.6) / (1.6762*10^-27)

Accerleration = 9.23 * 10^13

Therefore accerlation is 9.23 * 10^13 m/sec².

Answer 4

(4)

Distance between two plates = 5.00 cm = 0.05 m

Electric Field strength  = 606 v/m

E = V / d

V = E * d

Potential Difference = 606 * 0.05

Potential Difference = 30.3 V

Now if the Distance between two plates is increased to 7.00 cm = 0.07 m

Electric Field strength

E = V / d

E = 30.3 / 0.07

E = 432.85 V/m

Therefore the Electric field is decreased to 432.85 V/m when distance between the plates is 7.00 cm apart.

Answer 5

(5)

The mass of the small foam sphere = 2.70 * 10^-3 kg.

Distance between the Plates = 12.5 cm = 0.125 m.

Potential Difference applied = 5.05 * 10² V.

The ball moves right making angle = 0.613

Electric field = V / d = (5.05 * 10²) / 0.125 m = 4040 V/m.

Electric Field = 4040 V/m.

Now we need to calculate the force on the foam.

Figure showing the Free body diagram of Forces acting on the foam

From the figure,

T cosθ = mg

Fe = T sinθ

Where T is the tension

T Cos0.613 = (2.70 * 10^-3) * 9.8

T * 0.999 = 0.02646

T = 0.0242600 N

Force on the foam due to electric field Fe = Tsinθ

Fe = 0.02426 * sin0.613 = 2.8 * 10^-4 N

Force on the foam due to electric field  Fe = Charge of the foam * Electric field

Charge of the foam = Force on the Foam / Electric field

q = 7.007 * 10^-8 C.

Therefore the charge of the foam is 7.007 * 10^-8 C.