Question

A uniform electric field (constant in magnitude and direction) exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 cm distant from the first, in a time interval of 2.90×10−6 s .
a) Find the magnitude of the electric field. Express your answer with the appropriate units.
b) Find the speed of the proton when it strikes the negatively charged plate. Express your answer with the appropriate units.

Answer 1

Answer:

Part a)

$E = 39.7 N/C$

Part b)

$v_f = 1.10\times 10^4 m/s$

Explanation:

Part a)

Proton is released from one plate and strike at other plate in time

$t = 2.90 \times 10^{-6} s$

$d = 1.60 cm$

now we will have

$d = \frac{1}{2}at^2$

$0.0160 = \frac{1}{2}a(2.90 \times 10^{-6})^2$

$a = 3.8 \times 10^9 m/s^2$

now we know that

$a = \frac{qE}{m}$

$3.8 \times 10^9 = \frac{(1.6 \times 10^{-19})E}{(1.67 \times 10^{-27})}$

$E = 39.7 N/C$

Part b)

Speed of the proton when it strike the other plate can be calculated by kinematics equations

$v_f^2 - v_i^2 = 2ad$

$v_f^2 - 0 = 2(3.8 \times 10^9)(0.0160)$

$v_f = 1.10\times 10^4 m/s$