Potential Gradient [Question(s) regarding the concept]

AI Thread Summary
The discussion revolves around calculating the potential gradient of a solid metal sphere given its surface potential. The potential gradient is derived from the electric potential, which behaves like a point charge outside the sphere. Participants clarify that the gradient should be calculated as the derivative of the potential with respect to distance, rather than using average values between two points. There are multiple corrections regarding the charge value and the calculations leading to the potential gradient, with some confusion about the arithmetic involved. Ultimately, the correct approach emphasizes using the derivative to find the instantaneous gradient at the specified distance.
Antonius
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[Note from mentor: This was originally posted in a non-homework forum so it doesn't have the homework template.]

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Problem:
The surface of a solid metal sphere (radius r = 4.58 cm) is at potential V = 9,851 Volts. Find the magnitude of the potential gradient at distance 9.21r from the center of the sphere (outside the sphere), in V/m.
Assume: V(∞) = 0.


My point: I have this general understanding that the Potential Gradient equals to ## \frac { \Delta V } { \Delta x } ## where - ## x ## - is the distance between the charge ## Q ## and the point (or another charge.)

My question: Well in this problem, I am given ## V ## being 9851 Volts but that's on one side.
I watched an educational (physics) video saying that if there is only one charge and no other charges, then the Electric Potential Energy equals to ## zero. ## (which makes the Potential (at distance 9.21 * r = 0 V) That's being said, should I just divide the given ## V ## by the ## distance - x - ## which is also given. [ ## 9.21 * r ## ]?

Please help me with the concept. If you have a link to an online file that could explain this topic in depth please share it. I really want to learn but what I learned from my textbook is not leading me to find answers to problems I am encountering...
 
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DEvens said:
what the potential is at a radius from a particular charge
So I did this: $$ V = \frac {kq} {r}; 9851 Volts = \frac { 8.9 * 10^9 * q } { 0.0458 m } ; Hence, q = 5.069 C $$
Now since I know the ## q ## I decided to find ## V ## at ## 9.1 r ##.
To do so I followed these steps: $$ V = \frac { 8.9 * 10^9 * 5.069 } { 9.1 * 0.0458 } = 1.08 * 10^11 V $$
My question: Why is that the given Potential is smaller than the potential at ## 9.1 * r ## ?
Shouldn't it be smaller since ## V = \frac { U } { q } ## and ## U = \frac {k q} { r } ## ? :/
 
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Antonius said:
9851Volts=8.9∗109∗q0.0458m;Hence,q=5.069C


Double-check your arithmetic in this step.
 
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jtbell said:
Double-check your arithmetic in this step.
Apparently, I forgot to put a parenthesis while dividing by 8.9e9. jtbell, thank you!

So, q = 5.07e-7
Hence, ## \frac {\Delta V} {\Delta r} = -2.36e4 V/m ##

I am hoping this answer is correct.
 
Your q is still off by a factor of 10. Check the decimal points in your input numbers. I used the numbers in the first equation of post #3.
 
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jtbell said:
Your q is still off by a factor of 10. Check the decimal points in your input numbers. I used the numbers in the first equation of post #3.
jtbell said:
Your q is still off by a factor of 10. Check the decimal points in your input numbers. I used the numbers in the first equation of post #3.
Yes, I see. I wrote -7, instead of -8. The final answer that I have gotten is calculated with 5.069*10^(-8). Thank you!
 
Now for the second part (the gradient), how did you calculate that number, exactly? It's off by (approximately, not exactly) a factor of 10 also.
 
jtbell said:
Now for the second part (the gradient), how did you calculate that number, exactly? It's off by (approximately, not exactly) a factor of 10 also.
Hmm. What's going on lol.

Okay, let me work on it again.
so if q = 5.069 e -8

Now I need to find V at distance 9.1*r (away from center of sphere)
So V = kq/9.1*r = k*5.069 e -8 / (9.1*0.0458) = 1082.44 Volts

Potential Gradient = delta V / delta x = (1082.44 V - 9851 V) / (9.1*r - r) | 9.1*r - r = r ( 9.1 - 1) = 8.1*r

PG = - 23636.196 V/m = -2.36 e 4 V/m

Am I still wrong?
 
  • #10
I would call that the "average gradient" between the two r's. It's analogous to average velocity = (x2 - x1) / (t2 - t1).

The problem asks for the gradient at the second value of r (9.1r). This is analogous to the instantaneous velocity v = dx/dt. As DEvens noted in post #2, "gradient" here means "derivative with respect to r".
 
  • #11
jtbell said:
I would call that the "average gradient" between the two r's. It's analogous to average velocity = (x2 - x1) / (t2 - t1).

The problem asks for the gradient at the second value of r (9.1r). This is analogous to the instantaneous velocity v = dx/dt. As DEvens noted in post #2, "gradient" here means "derivative with respect to r".

If ## V (at 9.1r) = \frac {kq} {9.1*r} ## then ## \frac {dV} {dr} = \frac {kq} {9.1} *(\frac {-1} {r^2}) ## here ## q = \frac {V_r * r} {k} ## which I plugged into the equation and got ## \frac {dV} {dr} = \frac { -V_r} {9.1*r} ## which makes ## \frac {dV} {dr} = -23635.97 = -2.36e10^4 ##

What's wrong..
 
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  • #12
Take the derivative before substituting 9.1r (0.4168m) for r. This gives you an equation for the gradient at any r (provided that it's ≥ the radius of the sphere). Then substitute your desired value of r.
 
  • #13
jtbell said:
Take the derivative before substituting 9.1r (0.4168m) for r.
I did it before subbing a number for r though? And r is not 0.4168 m it's 0.0458 m ( in the question it says its r = 4.58 cm )
 
  • #14
When you take the derivative, you need to use an equation with a "generic" r, that can have any value. Your problem statement is confusing because it uses r and 9.1r to refer to "specific" values of r. I would have written it more carefully by calling the radius of the sphere r1 and the other radius r2 = 9.1r1.
 
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  • #15
jtbell said:
When you take the derivative, you need to use an equation with a "generic" r, that can have any value. Your problem statement is confusing because it uses r and 9.1r to refer to "specific" values of r. I would have written it more carefully by calling the radius of the sphere r1 and the other radius r2 = 9.1r1.
I got -2.15e5.

Is this what you also got?
 
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  • #16
That's it. :biggrin:
 
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  • #17
How did you guys get this result because I did it over and over with the values provided and I got something else. Here's a summary of the values used according to my calculations:

ro=0.0458 m
r=9.1r0= 0.41678 m
Q=5.02*10-8 C → from Vsurface
dV/dr=-kq/r2 V/m

⇒dV/dr= (8.99*109) N*m2/C2 * (5.02*10-8) C / (0.41678)2 m2

⇒dV/dr=-2597.36≅-2.60*103 V/m

 
  • #18
Alan I said:
How did you guys get this result because I did it over and over with the values provided and I got something else. Here's a summary of the values used according to my calculations:

ro=0.0458 m
r=9.1r0= 0.41678 m
Q=5.02*10-8 C → from Vsurface
dV/dr=-kq/r2 V/m

⇒dV/dr= (8.99*109) N*m2/C2 * (5.02*10-8) C / (0.41678)2 m2

⇒dV/dr=-2597.36≅-2.60*103 V/m

There is a vague step which I did not fully understand at first due to weak math background.

Use "the definition of derivative" and it will be much clear and you will most likely get a correct result
 
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