Question

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 1

`a=2.275*10^(17)m/s^(2)`

Step-by-step 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

`m_(e)a=|e|E\\`

the initial acceleration of electron is given by:

`a=(|e|E)/(m_(e) )`

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

`a=(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)`