Calculating Electric Force: Magnitude and Distance | Electric Field Problem

  • Thread starter Thread starter zelda1850
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
The electric field strength between two oppositely charged parallel plates is given as 2.0 x 10^3 Newtons per coulomb, with an electron located midway between them. To calculate the force exerted on the electron, the formula F = qE can be used, where q is the charge of the electron (-1.602 x 10^-19 C) and E is the electric field strength. The distance between the plates is not needed for this calculation since the electric field already provides the necessary information. By substituting the values into the equation, the magnitude of the force can be determined. This approach simplifies the problem by focusing on the relationship between electric field and force.
zelda1850
Messages
66
Reaction score
0

Homework Statement



the magnitude of the electric field strength between two oppositly charged parallel metal plate is 2.0 x 10 3 Newtons per coulomb. point p is located midway between the plates.

an electron is located at point p between the plates. caculate the magnitude of the force exerted by the electric field

Homework Equations



Felect = k • Q1 • Q2 / d2

k = 9.0 x 109 N•m2
q1 = 2.0 x 10 3
q2 = 2.0 x 10 3
d = missing

The Attempt at a Solution



how do i figure out the distance for this problem?
 
Physics news on Phys.org
If you've already got the electric field, you don't need a distance; you can just use the fact that F=qE. If your field exerts 2000 Newtons of force on every Coulomb of charge in it, and your particle's charge is -1.6 * 10^-19 Coulombs or whatever (this may not be accurate; I don't exactly know the elementary charge off the top of my head) then you can just plug those in... This help any?
 
Coloumbs law can be also be written as F = qE.

You are given for the value of the electric field strength and charge of the electron is known to be -1.602e-19 C. Now you can calculate for the electric force exerted on the charge by the field.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top