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Exercise:
An em electrostatic voltmeter is an electroscope that is used to measure a voltage. This is possible since an electroscope consists of two metal parts insulated against each other in other words a capacitor. Therefore the charge measured with an electroscope is proportional to the voltage across the electroscope. Two voltmeters with capacitance CaO and CbO are used to measure the voltage across a capacitor. The first one measures VaO and the second one VbO. Explain this phenomenon and calculate the capacitance of the capacitor and the correct voltage.

Solution:
When an electrostatic voltmeter is connected to the capacitor the resulting circuit corresponds to two capacitors em in parallel. The charge on the capacitor is distributed between voltmeter capacitance C_ and the capacitor capacitance C such that the voltage Delta V_ across both of them is the same but smaller than the original voltage Delta V_. The same happens with voltmeter capacitance C_ and voltage Delta V_. The charge can thus be expressed as Q C Delta V_ C+C_ Delta V_ C+C_ Delta V_ This can be solved for C: C+C_Delta V_ C+C_ Delta V_ CDelta V_-Delta V_ C_Delta V_-C_ Delta V_ C CF fracCbtimesVb-Catimes VaVa-Vb C approx resultCP- For the voltage across the capacitor alone without a voltmeter we find Delta V_ fracQC fracC+C_Delta V_C fracCF+C_Delta V_CF fracC_Delta V_-C_Delta V_+C_Delta V_-C_Delta V_Delta V_C_Delta V_-C_Delta V_ VF fracCb-CatimesVatimesVbCbtimes Vb-CatimesVa V approx resultVP
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Exercise:
An em electrostatic voltmeter is an electroscope that is used to measure a voltage. This is possible since an electroscope consists of two metal parts insulated against each other in other words a capacitor. Therefore the charge measured with an electroscope is proportional to the voltage across the electroscope. Two voltmeters with capacitance CaO and CbO are used to measure the voltage across a capacitor. The first one measures VaO and the second one VbO. Explain this phenomenon and calculate the capacitance of the capacitor and the correct voltage.

Solution:
When an electrostatic voltmeter is connected to the capacitor the resulting circuit corresponds to two capacitors em in parallel. The charge on the capacitor is distributed between voltmeter capacitance C_ and the capacitor capacitance C such that the voltage Delta V_ across both of them is the same but smaller than the original voltage Delta V_. The same happens with voltmeter capacitance C_ and voltage Delta V_. The charge can thus be expressed as Q C Delta V_ C+C_ Delta V_ C+C_ Delta V_ This can be solved for C: C+C_Delta V_ C+C_ Delta V_ CDelta V_-Delta V_ C_Delta V_-C_ Delta V_ C CF fracCbtimesVb-Catimes VaVa-Vb C approx resultCP- For the voltage across the capacitor alone without a voltmeter we find Delta V_ fracQC fracC+C_Delta V_C fracCF+C_Delta V_CF fracC_Delta V_-C_Delta V_+C_Delta V_-C_Delta V_Delta V_C_Delta V_-C_Delta V_ VF fracCb-CatimesVatimesVbCbtimes Vb-CatimesVa V approx resultVP
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Attributes & Decorations
Branches
Electrostatics
Tags
capacitance, capacitor, parallel circuit
Content image
Difficulty
(3, default)
Points
0 (default)
Language
ENG (English)
Type
Calculative / Quantity
Creator by
Decoration