You\'ve put 7 nC of charge on a 8 cm long, thin rod. The charge is approximately

Question

You've put 7 nC of charge on a 8 cm long, thin rod. The charge is approximately evenly spread out. You want to figure out how strong the electric field is 3 meters away from one of the ends of the rod along the axis defined by the rod's length. You decide to get an approximate answer by imagining that you can replace the rod with a point charge located at the position of the rod's center. How strong does this approximation estimate the field to be?  By how much does your approximation differ from the exact answer? Suppose that you hold a particle of charge 8 nC a distance 8 cm above the midpoint of the charged rod. How much electric force is exerted on the rod due to the point charge's electric field? What's the magnitude of the force exerted by the rod on the charged particle?

Explanation / Answer

Refer above figure,

For E by line charge density use equation,

E= (/40)*(1/b-1/(L+b))

Plugging given values,

E= (9*10^9*(7*10^-9/0.08))*(1/0.08-1/(3.08)) = 9588.1 N/C

For E by point charge use equation,

E=kq/r^2 = (9*10^9*7*10^-9)/3.04^2 = 6.817

There is a lot of difference between two fields.

When a charge q= 8nC is above the midpoint of the rod, electric field exerted by rod on charge

E=(L/40r)*(1/(sqrt(r^2+(L/2)^2)))

Where r= perpendicular distance and L = length

E= (9*10^9*7*10^-9)* (1/(sqrt(0.08^2+(0.08/2)^2))) =704.36 N/C

Force exerted by rod on the charge particle,

F=q*E= (8*10^-9*704.36) = 5.63*10^-6 N

By Newton’s third law equal magnitude force will be exert on the particle by rod.

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