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3 years ago

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An electron initially 3.00 m from a nonconducting infinite sheet of uniformly distributed charge is fired toward the sheet. The

electron has an initial speed of 420 m/s and travels along a line perpendicular to the sheet. When the electron has traveled 2.00 m , its velocity is instantaneously zero, and it then reverses its direction.
A- What is the surface charge density on the sheet?B- Given the same initial velocity, from what distance should the electron be fired if it is to just reach the sheet?

1 answer:

Nikolay [14]3 years ago

3 0

Answer:

θ = 4.438 * 10^-18 C / m^2

x = 2.0 m

Explanation:

Given:

V_1 = 420 m/s

Δ r = 2.0 m

m_e = 9.109 *10^-31 kg

q_e = 1.602 * 10^-19 C

ε_o = 8.85 * 10^-12

part a

Energy Principle:

E_k,1 - F_e*Δ r = 0

Electric Field strength of infinite sheet

E = θ / 2*ε_o

0.5*m_e* V_1^2 - E*q_e*Δ r = 0

0.5*m_e* V_1^2 - θ*q_e*Δ r / 2*ε_o = 0

θ = (m_e*ε_o* V_1^2) / q_e*Δ r

θ = (9.109*10^-31*8.85*10^-12* 420^2) / 1.602*10^-12*2

θ = 4.438 * 10^-18 C / m^2

part b

Since the Volt potential is constant of a infinite long uniform charge distribution:

0.5*m_e* V_1^2 - E*q_e*Δ r = 0

r_1 = x

r_2 = 0

0.5*m_e* V_1^2 - θ*q_e*x / 2*ε_o = 0

x = (m_e*ε_o* V_1^2) / q_e*θ

x = (9.109*10^-31*8.85*10^-12* 420^2) / 1.602*10^-12*4.438 * 10^-18

x = 2 . 0 m

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