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vsage
The electric potential along the x-axis (in kv) is plotted versus the value of x, (in meters). Evaluate the x-component of the electrical force (in Newtons)on a charge of 5.10 micro-C located on the x-axis at x=2.8 m.
http://www.geocities.com/vsage3/p.bmp
I tried finding the value of kV at x = 2.8 so I would have this:
dV/dx = E = F/q
q = 5.1e-06C
Hint: Use graphical techniques to evaluate the electric field, i.e., the x component of the electric field is the negative of the change of the potential with respect to x. Careful with units of potential (given in kV) and of charge (micro-C). In order to check the sign, remember in which direction the positive charge moves when located at the given position.
(this was given with the problem)
Any ideas? I have to get the value within 3% of what the computer says so that might be why I'm having such a hard time.
Edit: here is the value I got the time I tried it
V = Electric field * distance
V = Force / charge * distance
-2500V = F / 5.1e-06 * 2.8
F = -0.00455N but it's wrong according to the computer.
Edit thanks but I got the answer wrong too many times and I can't correct it. I heeded what you said but apparently I don't have a good enough grasp on the subject to apply it :(
http://www.geocities.com/vsage3/p.bmp
I tried finding the value of kV at x = 2.8 so I would have this:
dV/dx = E = F/q
q = 5.1e-06C
Hint: Use graphical techniques to evaluate the electric field, i.e., the x component of the electric field is the negative of the change of the potential with respect to x. Careful with units of potential (given in kV) and of charge (micro-C). In order to check the sign, remember in which direction the positive charge moves when located at the given position.
(this was given with the problem)
Any ideas? I have to get the value within 3% of what the computer says so that might be why I'm having such a hard time.
Edit: here is the value I got the time I tried it
V = Electric field * distance
V = Force / charge * distance
-2500V = F / 5.1e-06 * 2.8
F = -0.00455N but it's wrong according to the computer.
Edit thanks but I got the answer wrong too many times and I can't correct it. I heeded what you said but apparently I don't have a good enough grasp on the subject to apply it :(
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