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An electron and a proton are both released from rest, midway between the plates of a charged parallel-plate capacitor. The only force on each of the two particles is the force from the uniform electric field due to the capacitor. Each particle accelerates until striking one of the plates of the capacitor. (There is no gravity in this problem and we ignore the small force between the electron and the proton.) How do the final kinetic energies and final speeds (just before striking a plate) compare

Sagot :

Answer:

Explanation:

Let the potential difference between the middle point and one of the plate be ΔV .

electric potential energy will be lost and it will be converted into kinetic energy .

Electrical potential energy lost = Vq , where q is charge on charge particle .

For proton

ΔV× q = 1/2 M V² ( kinetic energy of proton )

where M is mass and V be final velocity of proton .

For electron

ΔV× q = 1/2 m v² ( kinetic energy of electron  )

where m is mass and v be final velocity of electron . Charges on proton and electron are same in magnitude .

As LHS of both the equation are same , RHS will also be same . That means the kinetic energy of both proton and electron will be same

1/2 M V² =  1/2 m v²

(V / v )² = ( m / M )

(V / v ) = √ ( m / M )

In other words , their velocities  are  inversely proportional to square root of their masses .

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