ashworcp
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Homework Statement
Use V=kq/r, E(x) = V/x, E(y) = V/y, E(z) = V/z to derive an expression for the electric field at a point charge q.
E(r) = ?
Homework Equations
E = F/Q
Deriving electric field from potential derivatives
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Use V=kq/r, E(x) = V/x, E(y) = V/y, E(z) = V/z to derive an expression for the electric field at a point charge q.
E(r) = ?
E = F/Q
Discussion
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Homework Helper
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I hope someone else will take a look at this!
I think it is simply false to say E(z) = V/z, assuming E(z) means the z component of the electric field. Consider the situation where the point (x,y,z) = (1,0,0) relative to the charge q causing the field.
Then E(z) = V/z = V/0 = infinity.
The electric field will not be infinity at a position 1 meter away from the charge.
Your E = F/Q equation allows you to find the formula for E.
It assumes there is another charge, Q at the position we are interested in and F = kq*Q/r². Use E = F/Q to finish the job. And note that for the (1,0,0) example, the z component is not infinite.
I think it is simply false to say E(z) = V/z, assuming E(z) means the z component of the electric field. Consider the situation where the point (x,y,z) = (1,0,0) relative to the charge q causing the field.
Then E(z) = V/z = V/0 = infinity.
The electric field will not be infinity at a position 1 meter away from the charge.
Your E = F/Q equation allows you to find the formula for E.
It assumes there is another charge, Q at the position we are interested in and F = kq*Q/r². Use E = F/Q to finish the job. And note that for the (1,0,0) example, the z component is not infinite.
ashworcp
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Wait are you saying its infinity?
ashworcp
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Those are the 3 given equations I'm suppose to use to create an expression for E(r)
Homework Helper
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Are you sure you have given the exact question?
By any chance would there be partial derivative symbols in those Ex, Ey and Ez formulas as you see here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/efromv.html#c1
By any chance would there be partial derivative symbols in those Ex, Ey and Ez formulas as you see here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/efromv.html#c1
ashworcp
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Yes, I am sorry I didn't know how to do that and forgot my negative sign.
Homework Helper
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Do you mean there ARE derivatives in those formulas?
So E(x) = ∂V/∂x, and so on?
By the way, you can copy and paste symbols like ∂ from this page:
https://www.physicsforums.com/blog.php?b=346 .
Okay, so what difficulty are you having in calculating E(x) = ∂V/∂x given V = kq/R, where R = (x² + y² + z²)½ ? Are you okay doing derivatives? When doing ∂V/∂x, you treat y and z like constants so the derivatives here are not difficult.
So E(x) = ∂V/∂x, and so on?
By the way, you can copy and paste symbols like ∂ from this page:
https://www.physicsforums.com/blog.php?b=346 .
Okay, so what difficulty are you having in calculating E(x) = ∂V/∂x given V = kq/R, where R = (x² + y² + z²)½ ? Are you okay doing derivatives? When doing ∂V/∂x, you treat y and z like constants so the derivatives here are not difficult.
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ashworcp
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So the derivative is R = 1/2(x^2+y^2+z^2)^-1/2 (2x+2y+2z)?
Homework Helper
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No. Do you have this?
V = kq/r = kq(x² + y² + z²)^-½
The derivative will not be R =, it will be Ex = or Ey = or Ez = something. You will have to do all three derivatives. Start with Ex = ∂V/∂x = ...
It is really a chain rule operation; if you are not an expert on that you may find it easier to let U = x² + y² + z², simplifying V to kqU^-½
Using the chain rule, you will differentiate U^-½ with respect to U and then multiply that by ∂U/∂x. In your R = 1/2(x^2+y^2+z^2)^-1/2 (2x+2y+2z) answer, you have done both parts incorrectly. In the first part you have basically said that the derivative of U^-½ with respect to U is U^-½, which is not correct. Look up the derivative of a power rule and check it! In the second part, you have to remember that y and z are constants because it is a partial derivative with respect to x.
Give it another try!
V = kq/r = kq(x² + y² + z²)^-½
The derivative will not be R =, it will be Ex = or Ey = or Ez = something. You will have to do all three derivatives. Start with Ex = ∂V/∂x = ...
It is really a chain rule operation; if you are not an expert on that you may find it easier to let U = x² + y² + z², simplifying V to kqU^-½
Using the chain rule, you will differentiate U^-½ with respect to U and then multiply that by ∂U/∂x. In your R = 1/2(x^2+y^2+z^2)^-1/2 (2x+2y+2z) answer, you have done both parts incorrectly. In the first part you have basically said that the derivative of U^-½ with respect to U is U^-½, which is not correct. Look up the derivative of a power rule and check it! In the second part, you have to remember that y and z are constants because it is a partial derivative with respect to x.
Give it another try!