Question

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 2.5 × 10-7 C/m2, and the plates are separated by a distance of 1.5 × 10-2 m. How fast is the electron moving just before it reaches the positive plate?

Answer 1

Speed = 1.22 x 10^(7) m/s

Step-by-step explanation:

Electric Field has a formula of;

E = σ/ε_o

Where,

σ is surface

εσ is vacuum permittivity

Also, acceleration is given by;

a = qE/m

Thus, replace E with σ/ε_o;

a = (qσ/ε_o)/m

Where;

q is charge of electron = 1.6 x 10^(-19) C

m is mass of electron = 9.11 x 10^(-31) kg

While ε_o has a value of 8.85 x 10^(-12) N.m²/C²

Thus, plugging in the relevant values to obtain ;

a = [((1.6 x 10^(-19))x 2.5 x 10^(-7)]/(8.85 x 10^(-12) x 9.11 x 10^(-31)) = 4.96 x 10^(15) m/s²

From Newton's 3rd law of motion,

V² = U² + 2as

Thus,plugging in the relevant values,

V² = 0² + 2(4.96 x 10^(15) x 1.5 x 10^(-2))

V² = 14.88 x 10^(13)

V = √14.88 x 10^(13) = 1.22 x 10^(7) m/s

Answer 2

Therefore, the velocity of the electron is 1.22 × 10⁷m/s

Step-by-step explanation:

The velocity of the electron can be found using the conservation of energy. The conservation of energy for the electron passing from one plate to another plate is,

`(1)/(2) mv^2=qV`

Expression for velocity

=
`v = \sqrt{(2q \sigma d)/(\varepsilon_0 m) }`

`v = \sqrt{(2(1.6*10^(-19)*2.5*10^(-7)(0.015))/((8.854*10^(-12)*(9.1*10^(-31)))) }`

`v =\sqrt{(1.2*10^(-27))/(8.05714*10^(-42)) } \\v = \sqrt{1.48936*10^(14)}`

v = 1.22 × 10⁷m/s

Therefore, the velocity of the electron is 1.22 × 10⁷m/s