Electromagnetism Question

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science_geeks
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#1
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#1
Hi,

I'm a bit stuck with this question. I have answered the first parts correctly but I'm really struggling to understand part c, calculating the charge on C2. please could someone help with this.

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Last edited by science_geeks; 2 years ago
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Missy_student.12
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#2
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(Original post by science_geeks)
Hi,

I'm a bit stuck with this question. I have answered the first parts correctly but I'm really struggling to understand part c, calculating the charge on C2. please could someone help with this.

Thanks in advance! Name:  Screenshot 2020-05-26 at 21.39.00.png
Views: 28
Size:  130.3 KBName:  IMG_6135.JPG
Views: 14
Size:  140.2 KBName:  IMG_6134.JPG
Views: 11
Size:  100.9 KB
Hi, I'm an A level student,

Are you GCSE student and what exam board are you if so?
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science_geeks
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(Original post by Missy_GCSE)
Hi, I'm an A level student,

Are you GCSE student and what exam board are you if so?
Hey, I'm an A-level physics student too and I'm doing AQA. Just finding physics really tricky this year!
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Missy_student.12
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(Original post by science_geeks)
Hey, I'm an A-level physics student too and I'm doing AQA. Just finding physics really tricky this year!
Ah ok. I understand. My friends are finding it a challenge too.

Unfortunately, I didn't take A level physics, and even if I did, our school does OCR A sciences.

Sorry , but I hope you do find answers to your questions.
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Joinedup
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first start to think of the 2μF and 3μF capacitors as being an effective capacitor of 5μF which is in series with the 20μF capacitor


Something interesting happens with capacitors in series - they all store the same charge as each other... here's an explanation of why

http://farside.ph.utexas.edu/teachin...es/node46.html

so the charge stored on the 5μF effective capacitor must be half of the total charge stored in the entire capacitor network - the other half is stored on the 20μF... and you can calculate the total charge using the value from part 2 and the 12V applied voltage

should be clear that the PD across the 5μF effective capacitor is different to the PD across the 20μF capacitor - the PD's across the real 2μF and 3μF is the same as the PD calculated for the 5μF effective capacitor.
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science_geeks
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(Original post by Missy_GCSE)
Ah ok. I understand. My friends are finding it a challenge too.

Unfortunately, I didn't take A level physics, and even if I did, our school does OCR A sciences.

Sorry , but I hope you do find answers to your questions.

No problem! It is tricky for certain but good luck to your friends!

Thank you
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science_geeks
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#7
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(Original post by Joinedup)
first start to think of the 2μF and 3μF capacitors as being an effective capacitor of 5μF which is in series with the 20μF capacitor


Something interesting happens with capacitors in series - they all store the same charge as each other... here's an explanation of why

http://farside.ph.utexas.edu/teachin...es/node46.html

so the charge stored on the 5μF effective capacitor must be half of the total charge stored in the entire capacitor network - the other half is stored on the 20μF... and you can calculate the total charge using the value from part 2 and the 12V applied voltage

should be clear that the PD across the 5μF effective capacitor is different to the PD across the 20μF capacitor - the PD's across the real 2μF and 3μF is the same as the PD calculated for the 5μF effective capacitor.
Ah I see! Thank you, that makes a lot more sense!
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