Electric field between two uniformly, linear charged rods

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SUMMARY

The discussion focuses on calculating the electric field strengths (E1, E2, E3) generated by two uniformly charged glass rods, each 10 cm long and charged to +10 nC, positioned 4.0 cm apart. The formula used for the electric field of a line charge is Eline = 1/(4πε0) * 2|λ|/r. The calculated values for E1 were initially estimated at 1.8E5 N/C, but the correct value reported by Mastering Physics is 1.25E5 N/C, highlighting the importance of considering the finite length of the rods in calculations.

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  • Understanding of electric fields and Coulomb's law
  • Familiarity with linear charge density (λ)
  • Knowledge of vector calculus for electric field calculations
  • Basic principles of electrostatics and finite line charge approximations
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  • Study the derivation and applications of Coulomb's law in vector form
  • Learn about electric fields produced by finite line charges
  • Explore the concept of linear charge density and its implications in electrostatics
  • Investigate the conditions under which approximations for electric fields are valid
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This discussion is beneficial for physics students, educators, and anyone interested in understanding electrostatics, particularly in calculating electric fields from charged objects.

tetris4life
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Homework Statement



Problem 27.9 Two 10-cm-long thin glass rods uniformly charged to +10nC are placed side by side, 4.0 cm apart.
What are the electric field strengths E1, E2, E3 and at distances 1.0 cm, 2.0 cm, and 3.0 cm to the right of the rod on the left, along the line connecting the midpoints of the two rods?

Homework Equations



Eline= 1/4πε0 * 2|λ|/r


The Attempt at a Solution



I understand that the Electric field at the midpoint would be zero. Point one and two will have the same strength but different directions.

I believe point one could be calculated by E1-(E3/3
Since the linear, uniform charge is inversely, linearly porportional.

For E1 I get 1.8E5 N/C; so, 1.8E5 N/C -(1.8E5N/C / 3) = 1.2E5 N/C

Mastering Physics reports the answer as 1.25E5 N/C
 
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Hello tetris4life,

Welcome to Physics Forums!

tetris4life said:

Homework Statement



Problem 27.9 Two 10-cm-long thin glass rods uniformly charged to +10nC are placed side by side, 4.0 cm apart.
What are the electric field strengths E1, E2, E3 and at distances 1.0 cm, 2.0 cm, and 3.0 cm to the right of the rod on the left, along the line connecting the midpoints of the two rods?

Homework Equations



Eline= 1/4πε0 * 2|λ|/r


The Attempt at a Solution



I understand that the Electric field at the midpoint would be zero. Point one and two will have the same strength but different directions.

I believe point one could be calculated by E1-(E3/3
Since the linear, uniform charge is inversely, linearly porportional.

For E1 I get 1.8E5 N/C; so, 1.8E5 N/C -(1.8E5N/C / 3) = 1.2E5 N/C

Mastering Physics reports the answer as 1.25E5 N/C

The equation which you used,
Eline= 1/(4πε0) * 2|λ|/r
is only valid for an infinitely long line (or infinitely long cylinder).

It also works as a very good approximation for the electric field near the midpoint of a finite line (or finite cylinder) if r << L, where r is the distance to the line and L is the length of the line. But in this problem, the distance to one of the rods is 3 cm, and the length of the rod is only 10 cm. It's true that 3 cm < 10 cm. But I would not go so far as to say 3 cm << 10 cm.

Starting with Coulomb's law (in vector form), if you go through the vector calculus to find the electric field, you will find that Mastering Physics is correct*.

*(correct in this particular instance, anyway.)
 

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