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Bookmarks People Window Help 99% 3) Mon 12:09 PM ucture.com'courses/1 1 1 84 1 2/assignments/688 1 7887module-item.id=1 1 6205 1 2 2018) Assignments>HW03 W03 (0%) Problem 5: Consider a parallel-plate capacitor made up of two conducting plates with dimensions 28 mm 23 mm platcs is.1.2.mm. what is the capacitancc, in pE.bewcen them storod on the positive plate, what is the potential, in volts, across the capaciter? 25% Part (b) if there is 088 nC o 25% Part (c) what is the magnitude of the electric 25% Part (d) tf the separation between the plates doubles, what will the electric field be if the charge is kept constant? field, in newtons per coulomb, inside this capacitor? Next Previous MacBook Pro 5

Explanation / Answer

a] Capacitance = Ae0/d = 28e-3 *23e-3*8.85e-12/1.2e-3 = 4.7495e-12 F

= 4.75 pF answer

b] Potential V = Q/C = 0.88e-9/4.75e-12 = 185.26 V

c] Electric field E = V/d = 185.26/1.2e-3 = 154386 N/C

d] It will be same as charge is same, E = 154386 N/C

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