Capacitance of a sphere

In summary, to calculate the total charge of a conducting sphere with a radius of 3400km and an electric field of 100 V/m at its surface, we use Gauss's Law to find the charge as 1.63*10^14 Coulombs. To calculate the potential at the surface using infinity as the reference point, we use the equation V(r) = -q/(4*pi*epsilon_0*r) and find it to be 4.07*10^28 J. To find the capacitance of the sphere using only one of the results from part a or b, we use the equation C = q/V and since capacitance is a geometric quantity, it is independent of both charge and potential, giving us
  • #1
Safder Aree
42
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?

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.
 
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  • #2
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
Safder Aree said:

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
ehild said:
You forgot ε0
You're right, it just slipped me while typing it up.
 
  • #5
kuruman said:
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.
 

1. What is capacitance of a sphere?

The capacitance of a sphere is a measure of its ability to store electrical charge. It is defined as the ratio of the charge on the sphere to the potential difference between the sphere and its surroundings.

2. How is capacitance of a sphere calculated?

The capacitance of a sphere can be calculated using the formula C = 4πε0r, where C is the capacitance, π is the mathematical constant pi, ε0 is the permittivity of free space, and r is the radius of the sphere.

3. What factors affect the capacitance of a sphere?

The capacitance of a sphere is affected by the radius of the sphere, the permittivity of the material surrounding the sphere, and the distance between the sphere and its surroundings.

4. How does capacitance of a sphere relate to its charge?

The capacitance of a sphere is directly proportional to its charge. This means that as the charge on the sphere increases, the capacitance also increases.

5. What are some real-life applications of capacitance of a sphere?

Capacitance of a sphere is used in many electronic devices, such as capacitors, which are used to store and regulate electrical charge. It is also used in electrical transmission lines and antennas.

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