Capacitor of a sphere and infinitely far hypothetical (spherical) plane

[dlslhc]

Homework Statement

An isolated sphere with charge +q can be treated as a capacitor whose second conductor is a hypothetical (spherical) plane with a ground potential and charge -q, located at a distance infinitely far from the sphere. the capacitance of the isolated conductive sphere is given by: C = 4pi Eo a, where C is the capacitance and a is the radius of the sphere. Calculate the charge q on a conductive sphere with a = 1.9 cm, when it is charged with a charging potential Vc = 6kV. Can you verify this calculation in the experiment? Explain briefly.

Homework Equations

The Attempt at a Solution

q = CV = 1.27 X 10^-8

No? because it is difficult to set up an infinitely far space without noise?

Thanks

 

[anathema]

for your question and attempt at a solution. Your calculation for the charge q on the conductive sphere with a radius of 1.9 cm and a charging potential of 6kV is correct. However, your concern about verifying this calculation in an experiment is valid.

In theory, the calculation is accurate and can be verified by measuring the charge on the sphere using a high-precision voltmeter. However, in practice, it may be difficult to set up an experiment to verify this calculation due to the challenges of creating a perfectly isolated and infinitely far space.

Additionally, there may be other factors that could affect the accuracy of the measurement, such as external electric fields or surface imperfections on the conductive sphere.

In order to accurately verify this calculation in an experiment, it would require careful control of all external factors and precise measurement techniques. Therefore, while the calculation is theoretically accurate, it may be challenging to replicate it exactly in an experiment.

 

[tomcue]

for the question! I can provide a response to this content by first acknowledging that the equation provided for the capacitance of an isolated conductive sphere is a commonly accepted formula. However, it is important to note that this is a simplified model and may not accurately represent real-world scenarios.

In order to calculate the charge on a conductive sphere with a radius of 1.9 cm and a charging potential of 6kV, we can use the given equation C = 4pi Eo a and substitute in the values to get C = 4 x 3.14 x 8.85 x 10^-12 x 1.9 cm = 2.25 x 10^-10 Farads. Then, we can use the equation q = CV to calculate the charge, which would result in q = (2.25 x 10^-10)(6000) = 1.35 x 10^-6 Coulombs.

However, as you mentioned, it may be difficult to verify this calculation in an experiment due to the challenges of creating an infinitely far space without any external noise or interference. In a real-world scenario, there may be other factors that can affect the capacitance and charge of the conductive sphere, such as the presence of other objects or varying environmental conditions.

Overall, while the given equation can provide an estimate of the charge on a conductive sphere in an idealized scenario, it may not accurately reflect the results in an actual experiment. As scientists, it is important to consider the limitations and assumptions of mathematical models and to validate our findings through experiments and observations.