Potential Difference problem with sphere and capacitor

In summary, a thin spherical shell made of plastic carries a uniformly distributed negative charge and is surrounded by two large thin disks made of glass carrying both positive and negative charges. The potential difference between the center of the plastic sphere and a point just outside of it is -202.32 V/m. The potential difference between a point just below the sphere and a point right beside the positive glass disk cannot be calculated without more information. It is important to pay attention to the signs when calculating potential differences.
  • #1
MAC5494
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Homework Statement



physics_problem.jpg


A thin spherical shell made of plastic carries a uniformly distributed negative charge -6e-10 coulombs (indicated as -Q1 in the diagram). Two large thin disks made of glass carry uniformly distributed positive and negative charges 1.5e-05 coulombs and -1.5e-05 coulombs (indicated as +Q2 and -Q2 in the figure). The radius R1 of the plastic spherical shell is 6 mm, and the radius R2 of the glass disks is 4 meters. The distance d from the center of the spherical shell to the positive disk is 18 mm.

(a) Find the potential difference V1 - V2. Point 1 is at the center of the plastic sphere, and point 2 is just outside the sphere.


(b) Find the potential difference V2 - V3. Point 2 is just below the sphere, and point 3 is right beside the positive glass disk.


Remember that the sign of the potential difference is important.


Homework Equations


Esphere = (1/4ε0)*(q/r2)
Ecapacitor = Q/A/ε0
ΔV = - (E*ΔL)


The Attempt at a Solution



a)
Ecapacitor = Q/A/ε0
= (1.5e-5)/(4^2∏)/(8.85e-12) = 3.4e4 N/C

ΔV = - (E*ΔL)
= - (3.4e4 * .006) = -202.32 V/m


b)

I don't understand how to do part b. I really could use some help. Part A is correct.
 
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  • #2
Please provide the figure.
 
  • #3
I added it.
 
  • #4
Is it unreasonable to think you can separately calculate the potential differences and add them up ?
(and take good care of the signs!)
 
  • #5
I've tried that, and it could be that I'm doing it wrong, maybe mixing up signs, but I'll take another look at it!
 
  • #6
Well, if we find something in the category "The attempt at a solution" in the next post, maybe we can help. That's what PFHW is for.
 
  • #7
Sorry, I meant to respond again, I figured out the problem. I was using the wrong equation for finding the potential difference due to the sphere.
 

FAQ: Potential Difference problem with sphere and capacitor

What is a potential difference problem involving a sphere and capacitor?

A potential difference problem with a sphere and capacitor involves calculating the potential difference (also known as voltage) between the two objects. This can be done by using the equation V = Q/C, where V is voltage, Q is charge, and C is capacitance.

How is capacitance related to potential difference?

Capacitance is a measure of an object's ability to store electric charge. In a potential difference problem with a sphere and capacitor, the capacitance of the capacitor is used in the equation to calculate the potential difference between the two objects.

What factors affect the potential difference between a sphere and capacitor?

The potential difference between a sphere and capacitor is affected by the distance between the two objects, the charge on the objects, and the material of the objects. A larger distance between the objects will result in a smaller potential difference, while a higher charge or a higher capacitance will result in a larger potential difference.

How can potential difference between a sphere and capacitor be measured?

Potential difference can be measured using a voltmeter. The voltmeter is connected to the two objects and will display the potential difference in units of volts (V).

What is the significance of solving a potential difference problem with a sphere and capacitor?

Solving a potential difference problem with a sphere and capacitor allows scientists to better understand and manipulate the behavior of electric fields and charges. This knowledge has many practical applications, such as in the design and function of electronic devices.

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