Total Charge on the Surface of a Drum

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SUMMARY

The discussion focuses on calculating the total charge on the surface of a drum using the formula Q=EAϵ0, where E is the electric field strength and A is the surface area. Initially, with a surface area of 6.25×10-2 m2, the charge was determined. When the surface area was increased to 0.124 m2, the same electric field of 1.30×105 N/C was required, necessitating a recalculation of the total charge. The permittivity of free space used in the calculations is 8.85×10-12 F/m.

PREREQUISITES
  • Understanding of electrostatics and electric fields
  • Familiarity with the concept of permittivity of free space
  • Ability to manipulate equations involving charge, electric field, and area
  • Basic knowledge of units in physics (e.g., square meters, coulombs)
NEXT STEPS
  • Research the implications of changing surface area on electric fields
  • Study the concept of electric field strength in different geometries
  • Learn about the applications of permittivity in various materials
  • Explore advanced electrostatic principles, such as Gauss's Law
USEFUL FOR

Students in physics, particularly those studying electrostatics, educators teaching electric field concepts, and anyone involved in practical applications of electric charge and field calculations.

y0tsubato
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Hi everyone-This is my first time posting a homework question. I'm a little stuck. I appreciate any help to clue me in the right direction.

Homework Statement



If the drum has a surface area of 6.25×10−2 , what total quantity of charge must reside on the surface of the drum?

I got this right by using Q=EA ϵ0


If the surface area of the drum is increased to 0.124 so that larger sheets of paper can be used, what total quantity of charge is required to produce the same 1.30×105 electric field just above the surface?


Homework Equations



permittivity of free space to be = 8.85×10−12
Q=EA ϵ0
electric field = 1.30*10^5
surface area= 6.25*10^-2


The Attempt at a Solution



(8.85*10^-12)(0.124)(6.25*10^-2)
I used the above because I thought it would be just a matter of multiplying the surface area and dividing by the electric field.
 
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The solution was very easy once I understood. My main problem was interpreting the phrase, "increased to". I kept trying to use the original surface area given.
 

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