Troubleshooting Electric Field Calculation: Qs, Rs, and N/C

In summary, the conversation is about calculating the electric field at a point between two charges, where the person is using a formula to do so. However, their answer differs from the answer key and they are seeking help in figuring out what they may be missing. The expert suggests using Coulomb's law and the superposition principle to calculate the net electric field at the point.
  • #1
joej
29
0
I'm doing the following problem, it seems to me that I am doing all the correct steps but my answer differs from the answer key, am I missing something?

1. What is the magnitude and the direction of the electric field at a point midway in between a -8.0uC and a +6.0uC change 4cm apart.


This is what I'm doing:

E = ( k * (Q1 / r1^2) ) * ( 1 + ( (Q2 / Q1) / (r2^2 / r1^2) ) )

==

(9x10^9 * ( (8 * 10^-6) / ( 2 * 10^-2) ) ) * ( 1 + ( (6 / 8) / (2 / 2) ) )

==

3600000 * 1.75 = 6300000 N/C


now... that is what I'm getting, could somoen please point out to me what I am forgetting to do, if anything.



__________edit

stupidity strikes again... forgot to square r1
 
Last edited:
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  • #2
Why write E in such a strange way? (to me anyway)

Use Coulomb's law and the superposition principle.
Calculate the field due to Q1 and the field due to Q2, then add.
 
  • #3
Hello joej...

The formula for electric field at a point distant r units from a charge Q is:

[tex]\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^2}\hat{e}_{r}[/tex]

The direction of the field is radial. In your problem you need to compute

[tex]\vec{E}_{net} = \vec{E_{1}} + \vec{E_{2}}[/tex]

the two terms arising due to the electric fields produced (independently--make a note of this, the superposition principle) has been used here) by the two charges. Have you used this fact? :confused:

Cheers
Vivek
 

Related to Troubleshooting Electric Field Calculation: Qs, Rs, and N/C

1. How do I calculate the electric field using Qs, Rs, and N/C?

In order to calculate the electric field using Qs (charge), Rs (distance), and N/C (electric field strength), you can use the formula E = Q/(4πε0Rs^2), where ε0 is the permittivity of free space. This formula can be used for point charges or for uniformly charged spheres.

2. What units should be used for Qs, Rs, and N/C in the electric field calculation?

Qs should be measured in Coulombs (C), Rs should be measured in meters (m), and N/C should be measured in Newtons per Coulomb (N/C) for the electric field calculation to work correctly. It is important to use consistent units throughout the calculation to avoid errors.

3. Why is the electric field calculation important in science and engineering?

The electric field calculation is important in science and engineering because it allows us to understand and predict the behavior of electric charges. This is crucial for designing and developing technologies such as electronics, power systems, and medical devices. It also helps us understand natural phenomena such as lightning and the Earth's magnetic field.

4. Can the electric field calculation be used for non-uniformly charged objects?

The basic formula for the electric field calculation can only be used for point charges or uniformly charged spheres. However, for non-uniformly charged objects, the electric field can be calculated by breaking the object into smaller, uniform sections and using the superposition principle to add up the contributions from each section.

5. What are some common sources of error in the electric field calculation?

Some common sources of error in the electric field calculation include using incorrect units, not considering the direction of the electric field (it is a vector quantity), and not accounting for the effects of other nearby charged objects. It is important to double check all inputs and assumptions to ensure an accurate calculation.

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