What Went Wrong with My Calculation for Problem 14?

  • Thread starter Thread starter mustang
  • Start date Start date
  • Tags Tags
    Coulomb's law Law
Click For Summary

Homework Help Overview

The discussion revolves around two physics problems involving electric fields and forces. The first problem concerns calculating the electric field between two charges, while the second problem focuses on determining the electric force and field strength related to an electron's acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the application of Coulomb's law for calculating electric fields and question the correct formulas for force and electric field strength. There are attempts to clarify the relationships between force, mass, and acceleration, as well as the implications of charge in these calculations.

Discussion Status

Some participants have provided guidance on relevant formulas and calculations, while others express uncertainty about their results and seek clarification on specific steps. Multiple interpretations of the problems are being explored, particularly regarding the calculations for problem 14.

Contextual Notes

There are indications of potential confusion regarding the values used in calculations, particularly in problem 14, where a participant questions their previous answer based on the mass of the electron and its acceleration.

mustang
Messages
169
Reaction score
0
Problem 11.
Find the magnitude electric field at a point midway between two charges of 34.4*10^9 C and 78.6*10^-9 C separated by a distance of 58.2 cm. Answer in N/C.
Note: Do i use coulomb's law? If so, when i multiply the constant to the quiotent is that my answer?

Problem 14.
An electron moving through an electric field experiences an acceleration of 6*10^3m/S^2.
a. Find the magnitude of the electric force acting on the electron. Answer in N.
Note: Do i use F=mass*acceleration?
b. What is the magnitude of the electric field strength? Answer in N/C.
What formula do I use?
 
Physics news on Phys.org
E=qF will help u
F=ma will do
 


Originally posted by mustang
Problem 11.
Find the magnitude electric field at a point midway between two charges of 34.4*10^-9 C and 78.6*10^-9 C separated by a distance of 58.2 cm. Answer in N/C.
Note: Do i use coulomb's law? If so, when i multiply the constant to the quiotent is that my answer?
Find the field from 1 then the field from the other then subtract.

field from first charge:
[tex]F = \frac{k q_1 q_2}{d^2}[/tex]

One of the charges isn't there so just divide it out.
[tex]\frac{F}{q_1} = \frac{k q_2}{d^2}[/tex]

[tex]\frac{F}{q_1} = \frac{(9x10^9)(34.4x10^-^9)}{0.291^2}[/tex]

[tex]\frac{F}{q_1} = 3656[/tex] N/C

field from second charge:
[tex]\frac{F}{q_1} = \frac{k q_2}{d^2}[/tex]

[tex]\frac{F}{q_1} = \frac{(9x10^9)(78.6x10^-^9)}{0.291^2}[/tex]

[tex]\frac{F}{q_1} = 8354[/tex] N/C

field at that point:
8354 - 3656 = 4698 N/C




Problem 14.
An electron moving through an electric field experiences an acceleration of 6*10^3m/S^2.
a. Find the magnitude of the electric force acting on the electron. Answer in N.
Note: Do i use F=mass*acceleration?
b. What is the magnitude of the electric field strength? Answer in N/C.
What formula do I use?

For question A:
You know the formula [tex]F = ma[/tex]. You know the mass of an electron and its rate of acceleration. Sub into find the force.

For question B:
The force is given by the field strength x charge. For field I'll just put L since I don't know what it should be.
F = Lq
L = F/q

You solved the force in part A and you know the charge of an electron.
 
Last edited:
Sorry

I rechecked and for problem 11 it was 34.4*10^-9C.
 
so the answer will be 4698 * 10^-9
 
Question for problem 14.

I had a=6*10^3 and m=9.109*10^-31. I subsituted those values for ma in F=ma and got 5.4654*10^-27. When i posted the answer I got it wrong. What did I do wrong?
 

Similar threads

Replies
3
Views
2K
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
6
Views
9K
Replies
4
Views
3K
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
6
Views
2K
Replies
2
Views
2K