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How to solve questions on electric field for Cambridge A level and UTME

Question 1

Two parallel metal plates, 4.0 cm apart, are at electric potentials of 800 V and 2000 V. Points X, Y and Z are situated in the space between the plates at distances of 1.0 cm, 2.0 cm and 3.0 cm from the lower plate

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What is the electric field strength, in V m–1, at X, Y and Z?

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{Cambridge A level may/june 2016 p12}

Solution

Electric field strenght at any point in two parallel plate is constant i.e it is a uniform electric field strenght

E = V/d

V is the potential difference between the plate = 2000 – 800 = 1200 v

d is the distance between the plate = 4.0 cm = 0.04 m

E = 1200 / 0.04 = 30000 vm-1 = 3 x 10^4 vm-1

C is the correct answer

Question 2

Two parallel vertical metal plates are connected to a power supply, as shown in Figure below

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An α-particle travels in a vacuum between the two plates. The electric field does work on the α-particle. The gain in kinetic energy of the α-particle is 15 keV. Calculate the electric field strength between the plates

{Cambridge A level 2016 may/june p22}

Solution

workdone = gain in kinetic energy

workdone = force x distance between the plates

workdone = qE x d

The charge on alpha particle = 2e

workdone = qEd = 3.2 x 10^-19 x E x 16 x 10^-3

wordone = 51.2 x 10^-22 E

1eV = 1.6 x 10^-19 J

1KeV = 1.6 x 10 ^-16 J

15 KeV = 24 x 10^-16 J

51.2 x 10^-22 E = 24 x 10^-16 J

E = 24 x 10^-16 / 51.2 x 10^-22

E = 0.46875 x 10 ^6 Vm-1

E = 4.7 x 10^5 Vm-1

Question 3

An oil droplet has charge –q and is situated between two horizontal metal plates as shown in the diagram below.

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The separation of the plates is d. The droplet is observed to be stationary when the upper plate is at potential +V and the lower plate is at potential –V.
For this to occur, what is the weight of the droplet?{Cambridge A level 2015 may/june p11}

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Solution

The droplet is stationary because;

the weight on the droplet = the electric force on the droplet

The electric force = qE

E = potential difference / distance between the plate

Potential difference = V – (-V) = 2V

E = 2V / d

Electric force = 2Vq / d

Weight on the droplet = 2Vq/d

B is the correct option

 

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