Chap_20_21_Disc[1]

# Chap_20_21_Disc[1] - 20-67: Find the electric energy...

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20-67 : Find the electric energy density between the plates of a 225- F parallel-plate capacitor. The potential difference between the plates is 345 V, and the plate separation is 0.223 mm. Picture the Problem : A capacitor stores energy when it is charged up to a certain voltage. Strategy: We can either find the total energy stored by the capacitor and divide it by the volume of the space between the plates, or we can apply equation 20-19, where the electric field between the plates is EVd (equation 20-4). The latter approach is utilized in the solution below, but both strategies would yield the same expression for E . u Solution: Combine equations 20-4 and 20-19 to find : E u  2 2 2 0 11 E0 0 22 2 2 12 2 2 3 2 3 2 8.85 10 C / N m 345 V 10.6 J/m 2 0.223 10 m V V uE dd       Insight: If we try the first suggested approach, we find   2 2 2 1 1 0 2 0 2 E 2 2 Ad V CV V u d , the same as above. Note that the energy density is independent of the plate area and therefore independent of the capacitance of the device. 21-32 : A 75-W light bulb operates on a potential difference of 95 V. Find (a) the current in the bulb and (b) the resistance of the bulb. (c) If this bulb is replaced with one whose resistance is half the value found in part (b), is its power rating greater than or less than 75-W? By what factor? Picture the Problem : A light bulb consumes electric power by drawing current at a certain voltage. Strategy: Use equation 21-4 to find the current drawn by the light bulb given its power rating and operating voltage. We can then use either equation 21-5 or 21-6 to find the resistance of the filament.

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## This note was uploaded on 10/13/2011 for the course PHY 102 taught by Professor Alexandrakis during the Fall '06 term at FIU.

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Chap_20_21_Disc[1] - 20-67: Find the electric energy...

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