Electric field, direction and magnitude

In summary: You can use the given formula for the electric field at the point due to a small element of the rod, but you have to integrate along the rod. In summary, the conversation is about determining the magnitude and direction of the electric field at a point along the axis of a uniformly charged rod. The attempted solution did not consider the direction of the electric field, leading to an incorrect value. The correct value is obtained by integrating along the rod using the given formula for the electric field due to a small element.
  • #1
minifhncc
46
0

Homework Statement



A rod 14.0 cm long is uniformly charged and has a total charge of -22.0μC. Determine
(a) the magnitude and
(b) direction of the electric field along the axis of the rod at a point 36.0 cm from its centre.

[I can do (b)]

Homework Equations



Electric field at the point due to a small element, ΔE=(k Δq)/(x^2)

The Attempt at a Solution


I got down to the total field at P being,
(kQ/l) * integral of 1/x^2 from x=0.29 and x=0.43

and I get 6.17 * 10^5 NC^-1

But the answer says 1.59 * 10^6 NC^-1

Are the answers wrong?
 
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  • #2
The given answer is right, how did you get that wrong value?

ehild
 
  • #3
ehild said:
how did you get that wrong value?

(kQ/l) * integral of 1/x^2 from x=0.29 and x=0.43

= 8.9876* 10^9 * -22 * 10^-6 / 0.14 ... etc
 
  • #4
minifhncc said:
(kQ/l) * integral of 1/x^2 from x=0.29 and x=0.43

= 8.9876* 10^9 * -22 * 10^-6 / 0.14 ... etc

And what is in place of etc?

ehild
 
  • #5
I don't think you have considered the direction of E.
E is a vector, and you need to treat the two components separately.
Check any book for the field due to a wire, and you'll understand what you've missed.
 
  • #6
It depends what is meant on "axis of the rod". It should be the straight line along the rod.

ehild
 

1. What is an electric field?

An electric field is a physical quantity that describes the strength and direction of the force that an electric charge would experience if placed in that field. It is represented by a vector, with both magnitude and direction.

2. How is the direction of an electric field determined?

The direction of an electric field is determined by the direction in which a positive test charge would move if placed in the field. It is always pointed in the direction of the force that a positive charge would experience.

3. What is the unit of measurement for electric field?

The unit of measurement for electric field is Newtons per Coulomb (N/C) in SI units. In other systems, it can also be measured in Volts per meter (V/m) or in CGS units, Statvolts per centimeter (statV/cm).

4. How is the magnitude of an electric field calculated?

The magnitude of an electric field can be calculated by dividing the force experienced by a test charge placed in the field by the magnitude of the test charge itself. Mathematically, it is represented as E = F/q, where E is the electric field, F is the force, and q is the test charge.

5. What are some real-world applications of electric fields?

Electric fields have a wide range of applications in our daily lives, such as in the functioning of electronic devices, power generation and transmission, and in medical equipment like MRI machines. They are also used in industrial processes, such as electroplating and electrostatic painting.

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