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Electric Field 2
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Electric Field 1
100

A proton has a ________ charge

positive charge

100

Give the formula for Work done. 

A - W = qEd

B - W = E/d

A

100

Give the SI unit for Magnetic flux

Weber (Wb)

100

Give the SI unit for Magnetic Field Strength

Tesla (T)

100

Give the SI unit for acceleration.

m/s/s

200

Give the charge on an electron.

A   (1.602 x 10^-19 C)

B  (-1.602 x 10^-19 C)

C (9.11 x 10^-31 C)

200

Calculate the work done on a proton (1.6 x 10^-19 C) which travels a distance of 20cm given the electric field strength is 2000 N/C.  



W = qEd

W = (1.6 x 10^-19) x 2000 x 0.2

W = 6.4 x 10^-17

200

Calculate the acceleration experienced by an electron travelling in a uniform electric field with a strength of 3.1 x 10^-5 N/C.

Given the calculated force

F = 4.97 x 10^-6 N

a = F/m

a = 4.97 x 10^-6 N/ 9.109 x 10^-31 kg

a = 5.5 x 10^6 m/s/s

200

Give the formula for torque

A - V x d

B - F x d

B

200

Calculate the acceleration experienced by an electron travelling in a uniform electric field with a strength of 3.1 x 10^-5 N/C. 

FIND THE FORCE

F = qE

F = (-1.602 x 10^-19) x (3.1 x 10^-5)

F = 4.97 x 10^-6 in the opposite direction

300

Give the mass of an electron

A - 1.67 x 10^-27 kg

B - 9.11 x 10^-31 kg 

C - 6.67 x 10^-31 kg

A

300

A uniform electric field is set up between two parallel plates that are 3mm apart with a potential difference of 12V.

What is the magnitude of the electric field?

E = V/d

E = 12/0.003

E = 4000 V/m

300

A student places a square coil of wire with a side length of 5cm into a uniform vertical magnetic field of 0.1T.

Calculate the magnetic flux

Ø = B x A

Ø = 0.1 x (0.05 x 0.05)

= 0.00025 Wb

= 2.5 x 10^-4 Wb

300

A motor lifts a mass of 400g. The motor has 200 turns in  0.3T magnetic field. If the radius is 0.142m

Determine the force required to lift the mass.

F = m x g

F = 0.4kg x 9.8m/s/s

F = 3.92N

300

How far apart are two conducting plates that have an electric field strength of 4.50× 103 V/m between them, if their potential difference is 15.0 kV?

d = V/E

d = V/E

d = 15 000/4500

d = 3.3m

400

Calculate the force on a proton given the electric field strength is 2000 N/C.



F = qE

F = (1.6 x 10^-19)(2000)

F = 3.2 x 10^-16

400

An electron travels with a velocity of 3 x 10^4 m/s at angle of 40o in a uniform electric field of magnitude 5000 N/C.

Find the acceleration of the electron due to the electric field.

a = Eq/m

a = 5000 (-1.6 x 10^-19) / (9.11 x 10^-31)

a = -8.8 x 10^14 m/s/s

400

A rectangular wire loop is connected to DC. One side of the loop is next to the magnet. The loop is free to rotate.

If the side length closest to the magnet is 20cm. Calculate to torque experienced by a 3kg mass.

Torque = F x d

Torque = mg x d

T = mg x d

T = 3 x 9.8m/s/s x 0.2

T = 5.88 N/m

400

Calculate the radius if the magnetic field strength is 0.020 T up the page. Each subatomic particle is fired with a velocity of 4000 m/s.

r electron = 

(9.11 x 10^-31)(4000) /(-1.6 x 10^-19)(0.02)

            =  1.1 x 10^-6 m

400

The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50 × 10^4 V/m.

What is the potential difference between the plates?

V = E x d

V = 7.5 x 104 x 0.04

V = 3000V

500

Calculate the electric field strength on a proton given the force is 200N.

E = F/q

E = 200/(1.6 x 10^-19)

E = 1.25 x 10^21 N/C

500

An electron travels with a velocity of 3 x 10^4 m/s at angle of 40o in a uniform electric field of magnitude 5000 N/C.

Find the acceleration of the electron due to the electric field.

a = Eq/m

a = 3000 (-1.6 x 10^-19) / (9.11 x 10^-31)

a = -5.28 x 10^14 m/s/s

500

A 5.0 cm wire carries a current of 0.75 A. Calculate the force acting on the wire when it is placed at right angles in a 0.60 T magnetic field.


F = ILB Sin (pheta)


F = 0.75 x 0.05 x 0.6 Sin 90

F = 0.0225 N   or 2.25 x 10-2N

500

A current of 0.87 A flows in a wire of length 1.4 m placed at 30o to a magnetic field of flux density 80 mT.

Calculate the force on the wire.

F = ILB Sin (pheta)

F = 0.87 x 1.4 x 0.08 Sin 30

F = 0.049 N   or 4.9 x 10-2N

500

The electric field strength between two parallel conducting plates separated by 5.00 cm is 7.50 × 104 V/m.

What is the potential difference between the plates?

V = E x d

V = 7.5 x 104 x 0.05

V = 3750V

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