Answered

An electron is released from rest at a perpendicular distance of 9.3 cm from a line of charge on a very long nonconducting rod. That charge is uniformly distributed, with 6.7 μC/m. What is the magnitude of the electron's initial acceleration?

Answer :

Khoso123

Answer:

[tex]a=2.275*10^{17}m/s^{2}[/tex]

Explanation:

The electron is negatively charged .The magnitude of force acting on on electron is

F=|e| E

Where E is electric field of long non conducting rod E=1/2πε₀r with linear charge density λ=6.7μC/m

From Newtons second laws

[tex]m_{e}a=|e|E\\[/tex]

the initial acceleration of electron is given by:

[tex]a=\frac{|e|E}{m_{e} }[/tex]

a=(|e|λ) /2πε₀rm= 2(|e|λ) / 4πε₀rm

Where 2 is multiplied on both numerator and denominator   to simplify calculation

The electron is initially at distance r=9.3cm=0.093 m from rod

So

[tex]a=\frac{2*8.99*10^{9}*1.6*10^{-19}*6.7*10^{-6} }{0.093*9.109*10^{-31} }\\ a=2.275*10^{17}m/s^{2}[/tex]

Other Questions