Total Charge on the Surface of a Drum

AI Thread Summary
To determine the total charge on the surface of a drum with a surface area of 6.25×10−2, the formula Q=EAϵ0 was applied, using the permittivity of free space as 8.85×10−12 and an electric field of 1.30×10^5. When the surface area is increased to 0.124, the same electric field requires recalculating the total charge. The initial confusion stemmed from misinterpreting the phrase "increased to," leading to the incorrect use of the original surface area. Once clarified, the calculation became straightforward, emphasizing the importance of correctly interpreting problem statements.
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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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