Equipotential surface / electric scalar potential problem (why )

I'm off to bed.In summary, the conversation discussed a potential field given by V = 3x^2*y - y*z and whether a statement about a unit normal to the equipotential surface at a specific point was valid. The individual attempted to solve the problem by filling in the given point and finding another point on the surface, but did not get the expected result. They then found the gradient of V and evaluated it at the given point, but made a calculation error. The conversation concluded with the individual realizing the mistake and understanding that the dot product of the gradient and the unit vector would be the gradient length if the unit vector was indeed a unit vector.
  • #1
FOIWATER
Gold Member
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equipotential surface / electric scalar potential problem (why!)

Homework Statement


A potential field is given by V = 3x^2*y - y*z. Is the following statement valid?

"A unit normal to the equipotential surface V = -8 at P(2,-1,4) is <-0.83,0.55,0.07>"

Homework Equations


Gradient of a scalar field?
dot product?

The Attempt at a Solution


First I filled in P(2,-1,4) to V to make sure P is on the equipotential surface.

(this point P is at the unit vector in question)

So I found another random point which also lies on the equipotential surface, I chose <0,1,8>

I then proceeded to apply dot product to both the unit vector in question, and the new vector I found, but did not get 0 (indicating they are not perp)

But I now believe the equipotential surface need not (is likely not, rather) flat in 3 space? is this assumption correct?

So I proceeded to take the gradient of V, and evaluate that at P. I received <-12,8,1>

I know, due to the definition of the gradient, that the direction of this vector indicates the direction of maximum increase of the scalar field V, which is normal to the equipotential surface at P? (would that make sense?)

I then proceeded to take the dot product of the unit vector in question, and the gradient evaluated at P but did not get 1 as I was expecting, rather, I got 14.43

I am assuming my mistake is assuming the gradient of V evaluated at P yields a vector which is perpindicular to V.

Can some one pls help

Tks

EDIT: I realized after I posted I should not of expected to get 1 with the dot product, but I should expect the product of the length of the two vectors. Also I do not get this
 
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  • #2
FOIWATER said:

Homework Statement


A potential field is given by V = 3x^2*y - y*z. Is the following statement valid?

"A unit normal to the equipotential surface V = -8 at P(2,-1,4) is <-0.83,0.55,0.07>"


Homework Equations


Gradient of a scalar field?
dot product?

The Attempt at a Solution


First I filled in P(2,-1,4) to V to make sure P is on the equipotential surface.

(this point P is at the unit vector in question)

So I found another random point which also lies on the equipotential surface, I chose <0,1,8>

I then proceeded to apply dot product to both the unit vector in question, and the new vector I found, but did not get 0 (indicating they are not perp)

But I now believe the equipotential surface need not (is likely not, rather) flat in 3 space? is this assumption correct?

So I proceeded to take the gradient of V, and evaluate that at P. I received <-12,8,1>

I know, due to the definition of the gradient, that the direction of this vector indicates the direction of maximum increase of the scalar field V, which is normal to the equipotential surface at P? (would that make sense?)

I then proceeded to take the dot product of the unit vector in question, and the gradient evaluated at P but did not get 1 as I was expecting, rather, I got 14.43

I am assuming my mistake is assuming the gradient of V evaluated at P yields a vector which is perpindicular to V.

Can some one pls help

Tks

EDIT: I realized after I posted I should not of expected to get 1 with the dot product, but I should expect the product of the length of the two vectors. Also I do not get this
You were perfectly correct in assuming that the gradient of V evaluated at P yields a vector which is perpendicular to the surface of constant V. But, it isn't a unit vector. However, it is pointing in the correct direction. To get the unit normal, you need to divide the gradient of V by its own magnitude.

Chet
 
  • #3
Thanks for the prompt reply,
I have a the gradient of V @ point P evaluated as <-12,8,1> and a unit vector in this direction as <-0.0574,0.03827,0.0042846>
Since this is not the unit normal given in the problem statement, I can assume this statement is false? (The textbook says it is a true statement)
 
  • #4
FOIWATER said:
Thanks for the prompt reply,
I have a the gradient of V @ point P evaluated as <-12,8,1> and a unit vector in this direction as <-0.0574,0.03827,0.0042846>
Since this is not the unit normal given in the problem statement, I can assume this statement is false? (The textbook says it is a true statement)

You have made a calculation error. Recheck your calculations.
 
  • #5
V=3x^2*y-yz
gradientV = <6xy,3x^2-z,-y>
@P(2,-1,4) gradientV = <-12,8,1>
length of gradient vector = 209
and unit vector is as above...
 
  • #6
aaaaaand i forgot to square root the sum of the squares
 
  • #7
oh, I got 14.43~!
which is what I got for the dot product of the gradient, and the unit vector given, earlier.
But I guess this makes sense, and I shouldn't be surprised, since knowing the unit vector in question WAS a unit vector, the dot product of the gradient and the unit vector would be the gradient length.

well, thanks, guys...
 

1. What is an equipotential surface?

An equipotential surface is a surface in space where the electric potential is the same at every point. This means that no work is required to move a charge from one point to another on the surface.

2. How do you determine the electric scalar potential?

The electric scalar potential, also known as the voltage, is determined by dividing the work done by the charge to move from one point to another by the magnitude of the charge. This can be calculated using the equation V = W/q, where V is the potential, W is the work, and q is the charge.

3. Why do we use equipotential surfaces in electric scalar potential problems?

Equipotential surfaces help us visualize and understand the electric field and potential in a given space. They also help us solve problems by providing a reference point where the potential is known to be constant.

4. How do you draw an equipotential surface?

To draw an equipotential surface, you need to know the configuration of the charges in the space and the values of the electric potential at different points. Then, you can plot points with equal potential values and connect them to form a surface.

5. What is the relationship between equipotential surfaces and electric fields?

Equipotential surfaces are always perpendicular to the electric field lines. This means that the direction of the electric field at any point on an equipotential surface is always tangent to the surface.

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