A 150 V battery is connected across two parallel metal plates of area 28.5cm 2 a

Question

A 150 V battery is connected across two parallel metal plates of area 28.5cm2 and separation 8.20mm, which form a capacitor. (a) Find its capacitance. (b) Find the charge stored on each plate, the electric field between the plates and the energy within the capacitor. (c) An electron is released from rest near the negative plate. What is its speed when it reaches the positive plate? (d) While the battery is still connected, the plates are pushed together until their separation is halved. Find the charge stored on each plate, the electric field between the plates and the energy within the capacitor.

Explanation / Answer

Accelerating Voltage = V1=1.60 kV = 1600 V

Charge on alpha= q=3.2*10^-19 C

Mass = m =6.64*10^-27 kg

Kinetic energy = electric potential energy

(1/2)mv^2 = qV

v = sq rt [2qV/m]

The alpha particles will emerge undeflected from between the plates if magnetic force=elctric force

qvB =qE

E = Vo/d

B = E /v

B =(Vo/d) sq rt [m/2qV]

B =0.04968 T

B =4.968*10^-2 T

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