Electric field between two uniformly, linear charged rods

In summary, the problem involves two uniformly charged glass rods placed 4 cm apart, and the question asks for the electric field strengths at distances 1 cm, 2 cm, and 3 cm to the right of the left rod along the line connecting the midpoints of the two rods. While an approximate equation can be used for the electric field near the midpoint of a finite line, it is not accurate enough for this problem where the distance is not much smaller than the length of the rod. Using Coulomb's law and vector calculus, the correct answer is 1.25E5 N/C.
  • #1
tetris4life
1
0

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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  • #2
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.)
 

FAQ: Electric field between two uniformly, linear charged rods

What is an electric field?

The electric field is a physical quantity that describes the force experienced by a charged object in an electric field. It is represented by a vector that shows the direction and magnitude of the force.

How is the electric field between two charged rods calculated?

The electric field between two uniformly charged linear rods can be calculated using Coulomb's Law. This law states that the electric field is directly proportional to the product of the charges on the rods and inversely proportional to the square of the distance between them.

What factors affect the strength of the electric field between two charged rods?

The strength of the electric field between two charged rods is affected by the magnitude of the charges on the rods, the distance between them, and the medium surrounding the charged rods. The electric field is stronger when the charges are larger and closer together, and weaker in a medium with higher permittivity.

Can the electric field between two charged rods ever be zero?

Yes, the electric field between two charged rods can be zero at certain points in space. These points, known as equilibrium points, occur when the forces from the two rods cancel each other out. This can happen when the charges on the rods are equal and opposite, or when the rods are placed at specific distances from each other.

How is the direction of the electric field between two charged rods determined?

The direction of the electric field is determined by the direction of the force that a positive test charge would experience in the electric field. The direction of the electric field is always from the positively charged rod towards the negatively charged rod.

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