# Electric fields

1. Nov 19, 2007

### pinkyjoshi65

what are the magnitude and direction of the electrical field strength at point Z in the situation below?
negatively charged sphere, x(-2.0*10-5C) is left most, 60 cm to the right of it is a positively charged sphere,Y (8.0*10-6C). Point Z is 30 cm to the right of Y.

What I did:
First i found E1, where q1= -2*10-5 C and r= 0.90 m (direction is right)

Then I found E2, where q2= 8.0*10-6 C and r= 0.30m (direction is left)
then i found the vector sum of E1 and E2, and found the field strenth at Z (direction is Left)

Is this right??

2. Nov 19, 2007

### Sourabh N

3. Nov 19, 2007

### pinkyjoshi65

final answer is 577800 N/C (left)

4. Nov 19, 2007

### pinkyjoshi65

got another question--not a problem..if a stationary charged test particle is free to move in an electrical field, in what direction will it begin to move?

5. Nov 19, 2007

### pinkyjoshi65

Anyone..??

6. Nov 19, 2007

### Galileo's Ghost

That depends on the direction of the field and the nature of the test charge. Typically, a "test" charge is a positive and will be accelerated in the same direction as the field.

7. Nov 19, 2007

### pinkyjoshi65

ah..k..i see..i saw in one of the websites, that the charge moves in a circular direction..

8. Nov 19, 2007

### Galileo's Ghost

9. Nov 19, 2007

### pinkyjoshi65

Last edited by a moderator: Apr 23, 2017
10. Nov 19, 2007

### Galileo's Ghost

The simulation I see when I click that link shows both an electric and magnetic field. The charged particle in question also has an initial velocity...Unless you are setting the parameters to some other situation...

11. Nov 19, 2007

### pinkyjoshi65

k i guess then ur right..one more question:three small, negatively charged spheres are located at the vertices of an equilateral triangle.The magnitudes of the charges are equal. Sketch the electical field in the region around this charge distribution, including the space inside the triangle.

ok..so all i know is that density of field lines will be more whr the field is greater..