Capacitance of a sphere

  • #1
Safder Aree
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Homework Statement


Assume a conducting sphere has a radius of 3400km with an electric field of 100 V/m at it's surface.
a) Calculate total charge of sphere.
b)Calculate potential at the surface using infinity at reference point
c) Calculate capacitance of the sphere using the result of a or b but not both.

I know how to do all the problems but how do I go about doing part C while using only one of the results?



Homework Equations



-Gauss's Law
-Potential
-Capacitance


The Attempt at a Solution


a)
$$\oint E.dA = \frac{q}{\epsilon_0}$$
$$q = E(4\pi*R^2)\epsilon_0$$
$$q=100*4\pi*(3.6*10^5)$$
$$1.63*10^{14} Coulombs$$

b)
$$V(r) = -\int E.dl$$
$$= -\frac{q}{4\pi\epsilon_0} \int_\infty^R \frac{1}{r^2}dr$$
$$=\frac{q}{4\pi\epsilon_0 R}$$
$$=\frac{1.63*10^{14}}{4\pi\epsilon_0(3.6*10^5)}$$
$$4.07*10^{28} J$$

c) $$C = \frac{Q}{\Delta V}$$.
I have no idea how to appraoch it using only variable. I know the answer must be Q/V from a and b.
Thank you for any guidance.
 
Last edited:

Answers and Replies

  • #2
kuruman
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You will see it immediately if you stop thinking in terms of numbers and start thinking in terms of symbols. You have ##V = \frac{q}{4 \pi \epsilon_0r}## and ##V = \frac{q}{C}##. Put it together. Capacitance is a geometric quantity and depends neither on ##q## nor on ##V##.
 
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  • #3
ehild
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The Attempt at a Solution


a)
$$\oint E.dA = \frac{q}{\epsilon_0}$$
$$q = E(4\pi*R^2)$$
You forgot ε0
 
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  • #4
Safder Aree
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You forgot ε0
You're right, it just slipped me while typing it up.
 
  • #5
Safder Aree
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You will see it immediately if you stop thinking in terms of numbers and start thinking in terms of symbols. You have ##V = \frac{q}{4 \pi \epsilon_0r}## and ##V = \frac{q}{C}##. Put it together. Capacitance is a geometric quantity and depends neither on ##q## nor on ##V##.

I understand now, thank you for the clarification.
 

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