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Capacitance of a sphere

1. 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?



2. Homework Equations

-Gauss's Law
-Potential
-Capacitance


3. 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

kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
8,327
1,871
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##.
 
ehild
Homework Helper
15,361
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3. The Attempt at a Solution
a)
$$\oint E.dA = \frac{q}{\epsilon_0}$$
$$q = E(4\pi*R^2)$$
You forgot ε0
 
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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