If an electron is accelerated through a potential difference of 500 V between two parallel plates separated by a distance of 2.0 cm. The change in kinetic energy of the electron during this motion is.

8.01×10^-17 J

Explanation: potential difference (v)= 500v, distance between plates (d) = 2cm = 0.02m

q = magnitude of an electronic charge = 1.602×10^-19c

E = strength of electric field = ?

Relationship between potential difference, disntahge between plates and electric field strength is given by the formule below

v =Ed

500 = E × 0.02

E = 500/ 0.02

E =25,000 v/m

But F=Eq

F = 25,000 × 1.602×10^-19

F = 4.005×10^-15 N

Using the work-energy theorem, the work done in moving the electron between the plates equals the kinetic energy of the electron.

Kinetic energy = work done = f × d

Kinetic energy = 4.005×10^-15 × 0.02

Kinetic energy = 8.01×10^-17 J

The change in kinetic energy of the electron is 8.0 x 10⁻¹⁷ J

Explanation:

For an electron to be accelerated through a potential difference in an electric field, work must be done on that electron to accelerate it.

When work is done on the electron, it gains energy and accelerates.

The work done to accelerate the electron is equal but opposite to the kinetic energy gained by the electron

                W = - ΔE     where W is work done

                                   ΔE is change in energy of the electron

The kinetic energy gained by the electron is measured in Electron Volt (eV), which is the amount of work required to move an electron through a potential difference of 1 Volts.

Hence 1 eV = energy gained by an electron when it is accelerated through a potential difference of 1 volts.

Therefore 500 eV would be the energy gained by the electron when accelerated through a potential difference of 500 V.

Using Coulombs equation which states that

                    E = qV   where E is the change in kinetic energy

                                            q is the electric charge

                                            V is the electron volt

Therefore,

                 E = (1) x (500) x 1.6 x 10⁻¹⁹J  

                                                       (1.6 x 10⁻¹⁹ is the quantity of charge in 1eV)

                    E= 8.0 x 10⁻¹⁷ J

answer: 0.20

explanation: this can only be the answew because all the other measuremnets are not close

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete