How Do You Calculate the Capacitance of a Sphere Using Only Charge or Potential?

[Safder Aree]

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.

 

[kuruman]

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]

The Attempt at a Solution

a)
$$\oint E.dA = \frac{q}{\epsilon_0}$$
$$q = E(4\pi*R^2)$$

You forgot ε₀

 

[Safder Aree]

You forgot ε₀

You're right, it just slipped me while typing it up.

 

[Safder Aree]

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.