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

Answer:

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

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,

$\frac{1}{2} mv^2=qV$

Expression for velocity

= $v = \sqrt{\frac{2q \sigma d}{\varepsilon_0 m} }$

$v = \sqrt{\frac{2(1.6\times10^{-19}\times2.5\times10^{-7}(0.015)}{(8.854\times10^{-12}\times(9.1\times10^{-31}))} }$

$v =\sqrt{\frac{1.2\times10^{-27}}{8.05714\times10^{-42}} } \\v = \sqrt{1.48936\times10^{14}}$

v = 1.22 × 10⁷m/s

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

Answer 2

Answer:

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

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