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Co= V= Part (a) Enter an expression for the capacitance, Co. of this parallel-plate capacitor in terms of the quantities given in the problem statement. 24= 20 b h k P €0 b I m R 20 b h k P Ho d I Ho E j m R n S HO d I Part (d) Eliminate Co and Vo from the expression in the previous step using the results of the earlier steps. Input the resulting expression for U. U= 1/2 (e A/d) (Ed)² ✓ Correct! Part (e) Enter an expression for the volume of the capacitor in terms of the parameters given in the problem statement. m R A A h k P e X J n S A e J () 7 8 9 TL 4 5 6 1 2 3 0 n S Part (1) The energy stored in the capacitor is represented by an upper-case U while the energy density is represented by a lower-case u. Divide the expression from part (d) by the volume of the parallel-plate capacitor to obtain the energy density, and enter a simplified expression for u. (The result obtained generally applies to the energy density stored in an electric field even though we just derived it specifically in the context of a parallel-plate capacitor) +- END NO BACKSPACE CLEAR DEL 789 TAAL 4 5 6 1 23 HOME - (789 HOME TL 4 5 6 1 2 3 + 0 NO BACKSPACE DEL 1 + 0 END NO BACKSPACE DEL CLEAR END CLEAR HOME

Co= V= Part (a) Enter an expression for the capacitance, Co. of this parallel-plate capacitor in terms of the quantities given in the problem statement. 24= 20 b h k P €0 b I m R 20 b h k P Ho d I Ho E j m R n S HO d I Part (d) Eliminate Co and Vo from the expression in the previous step using the results of the earlier steps. Input the resulting expression for U. U= 1/2 (e A/d) (Ed)² ✓ Correct! Part (e) Enter an expression for the volume of the capacitor in terms of the parameters given in the problem statement. m R A A h k P e X J n S A e J () 7 8 9 TL 4 5 6 1 2 3 0 n S Part (1) The energy stored in the capacitor is represented by an upper-case U while the energy density is represented by a lower-case u. Divide the expression from part (d) by the volume of the parallel-plate capacitor to obtain the energy density, and enter a simplified expression for u. (The result obtained generally applies to the energy density stored in an electric field even though we just derived it specifically in the context of a parallel-plate capacitor) +- END NO BACKSPACE CLEAR DEL 789 TAAL 4 5 6 1 23 HOME - (789 HOME TL 4 5 6 1 2 3 + 0 NO BACKSPACE DEL 1 + 0 END NO BACKSPACE DEL CLEAR END CLEAR HOME

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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I'm needing help on parts a, e, and f, please

Problem 3: An air-filled parallel-plate capacitor has plates with area A separated by a gap of length d resulting in a capacitance value Co. The plates were connected to a battery, and the capacitor was charged until the potential difference between the plates reached Vo, and finally the battery was removed. Co= V= 24= Part (a) Enter an expression for the capacitance, Co. of this parallel-plate capacitor in terms of the quantities given in the problem statement. 20 b h k P €0 b I m R HO d I 20 b h k P m R Ho E j n S Part (d) Eliminate Co and Vo from the expression in the previous step using the results of the earlier steps. Input the resulting expression for U. U= 1/2 ( A/d) (Ed)² ✓ Correct! но d I Part (e) Enter an expression for the volume of the capacitor in terms of the parameters given in the problem statement. m R A e j A h k P n S X A e (D) 789 HOME TL 4 5 6 3 1 2 0 NO BACKSPACE CLEAR + - END j n S 7 8 9 T^^ 4 5 6 1 2 3 + 0 NO BACKSPACE - Part (f) The energy stored in the capacitor is represented by an upper-case U while the energy density is represented by a lower-case u. Divide the expression from part (d) by the volume of the parallel-plate capacitor to obtain the energy density, and enter a simplified expression for u. (The result obtained generally applies to the energy density stored in an electric field even though we just derived it specifically in the context of a parallel-plate capacitor) - (7 8 9 HOME TAAL 4 5 6 /12 3 + 1 → + END 0 VO BACKSPACE CLEAR di HOME A END CLEAR
Transcribed Image Text:Problem 3: An air-filled parallel-plate capacitor has plates with area A separated by a gap of length d resulting in a capacitance value Co. The plates were connected to a battery, and the capacitor was charged until the potential difference between the plates reached Vo, and finally the battery was removed. Co= V= 24= Part (a) Enter an expression for the capacitance, Co. of this parallel-plate capacitor in terms of the quantities given in the problem statement. 20 b h k P €0 b I m R HO d I 20 b h k P m R Ho E j n S Part (d) Eliminate Co and Vo from the expression in the previous step using the results of the earlier steps. Input the resulting expression for U. U= 1/2 ( A/d) (Ed)² ✓ Correct! но d I Part (e) Enter an expression for the volume of the capacitor in terms of the parameters given in the problem statement. m R A e j A h k P n S X A e (D) 789 HOME TL 4 5 6 3 1 2 0 NO BACKSPACE CLEAR + - END j n S 7 8 9 T^^ 4 5 6 1 2 3 + 0 NO BACKSPACE - Part (f) The energy stored in the capacitor is represented by an upper-case U while the energy density is represented by a lower-case u. Divide the expression from part (d) by the volume of the parallel-plate capacitor to obtain the energy density, and enter a simplified expression for u. (The result obtained generally applies to the energy density stored in an electric field even though we just derived it specifically in the context of a parallel-plate capacitor) - (7 8 9 HOME TAAL 4 5 6 /12 3 + 1 → + END 0 VO BACKSPACE CLEAR di HOME A END CLEAR
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