Question

An electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The initial speed of the electron is 5.69 x 106 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

Answer 1

magnitude is 1382.59 N/C

Step-by-step explanation:

Given the data in the question;

The time taken is;

t = x / v

we substitute;

t = ( 2 × 10⁻²) / ( 5.69 × 10⁶ )

t = 3.5149 × 10⁻⁹ s

next, the acceleration is;

a = 2y/t² = [2( 0.150 × 10⁻²)] / [ ( 3.5149 × 10⁻⁹ )² ]

a = 2.42826 × 10¹⁴ m/s²

now, the electric field is;

E = ma / q

we know that;

mass of electron m = 9.11 × 10⁻³¹ kg,

charge of electron q = 1.60 × 10⁻¹⁹ coulomb

we substitute

E = ( 9.11 × 10⁻³¹ )(2.42826 × 10¹⁴) / 1.60 × 10⁻¹⁹

E = 2.21214 × 10⁻¹⁶ / 1.60 × 10⁻¹⁹

E = 1.3826 × 10²¹

E = 1382.59 N/C

Therefore, magnitude is 1382.59 N/C