Electric capacitance adding charged particals, rate

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SUMMARY

The discussion focuses on calculating the electric capacitance of the Earth and the impact of charged cosmic ray particles on its electrostatic potential. The capacitance is determined using the formula C = 4 * π * ε * R, yielding a value of 709.53 mF based on Earth's radius of 6380 km. Additionally, the rate of cosmic ray particles arriving at Earth is approximately 1 particle per square centimeter per second, which raises questions about the current and the electrostatic potential's rate of increase. The relationship between charge flow and electrostatic potential is explored, emphasizing the need for a deeper understanding of charge dynamics.

PREREQUISITES
  • Understanding of electric capacitance and its calculation methods
  • Familiarity with cosmic ray physics and particle charge
  • Knowledge of current measurement in Amperes and Coulombs
  • Basic principles of electrostatics, including potential and charge dynamics
NEXT STEPS
  • Research the derivation of the capacitance formula for spherical conductors
  • Learn about cosmic ray detection methods and their implications for Earth’s charge
  • Study the relationship between current, charge, and time in electrostatics
  • Explore the concept of divergence in electric fields and its applications
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Physics students, electrical engineers, and researchers interested in electrostatics, cosmic ray interactions, and the fundamental principles of capacitance.

cheff3r
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Okay so the question is

15. (a) Calculate the electric capacitance of the Earth.
(b) Positively charged cosmic ray particles are known to arrive at Earth at the rate of
approximately 1 particle per square centimetre per second. If each particle is
assumed to be singly charged, calculate:-
(i) the electric current carried to the Earth by these particles;
(ii) the rate at which the electrostatic potential of the Earth would rise if there
were no other charge movements to or from the Earth.

my solution
a) knowing Earth's radius of 6380km
C=4*(pi)*ε*R
=4*3.141*8.85*10^-12*6380*10^3
=709.53 mF

b) my problem with b is up until now the queestions not involving flow of charge, so here's my guess its something to do with divergence (we slightly touched on the topic in class) on the other hand doesn't it take more energy everytime charge is added to a capacitor so wouldn't that mean it is dependent on time?
 
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(b.i) Current is measured in Amperes. 1 Ampere = 1 Coulomb/second. How many Coulombs/second total reach the Earth's surface?

(b.ii) You know from part (a) that the potential of the Earth is
V = \frac{Q}{4\pi\epsilon_{0}R}
where Q s the total charge on the Earth at a given time. Can you find an expression for \frac{dV}{dt} ?
 

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