A potential difference of 1200 V is established between two parallel plates of a capacitor The plates of the - Physics - Electrostatic Potential And Capacitance

Question

A potential difference of 1200 V is established between two
parallel plates of a capacitor The plates of the capacitor are at a
distance of 2cm apart An electron is released from the negative
plate at the same instant a proton is released from the positive
end How does the following:
a) velocity
b) Energy compare when they strike the opposite plates
c) How far from the positive plate will the pass each other?
ANS: a) 42 84, b) equal, c)2 7cm

Ravi G · Accepted Answer

Voltage V = 1200 vDistance between the plates d = 2 cm= 0
02 mElectric field between the plates = Vd = 12000
02 = 6×104 v/ma) force on the particle in electric field isF = qE = mawhere m is the mass of the particle and a is the acceleratio of the particle
a = qEmon substituting the values for proton and electron we can calculate the value of acceleration for both
As distance travelled by them will be equal to the distance between the two plates
d = ut + 12at2as initially both are at rest
u = 0d =  12at2t = 2dat = 2dmqEVelocity with which they strike to the plate is :v = u + atv = 2dmqE×qEm = 2qEdm2)As energy is given by: U = qVcharge oth the partcile is same
 so energy with which they strike to the plate will be the same
3)Let x be the distancetravelled by proton from positive plat in time tx = ut + 12apt2x = 0 + 12apt2x = 12apt2Distance travelled by electron in same time t is
0
02 - x =12aet2We know the acceleration for bothe the particles, so we can solve these two equations and can  calculate the value of x