You place a point charges, of + 5 00 mu C at y = 20.0 cm and a point charge of -

Question

You place a point charges, of + 5 00 mu C at y = 20.0 cm and a point charge of -5 00 mu C at y=-20.0 cm. What is the electric potential at x = 0. Y = 0? How much work was done to bring the charges to their current locations, beginning at Infinite separation? Is the work + or - Consider a capacitor composed of 1 cm^2 plates separated by a layer of mica 0.02 cm thick. You connect the capacitors to a 90 V battery. What would the electric field in the region of space between the plates be if the plates were separated by vacuum? Suppose the dielectric constant for mica to be 6. What then is the electric field then between the plates? What is the capacitance of this mica filled capacitor? Now consider the global capacitance, a sphere of 4000 km radius, at a distance 50 km from ground potential. If the electric field in the atmosphere is 100 Volts/m, pointing downward, and E * 4 pi R^2 = Q/epsilon _ 0, what is Q/4 pi R^2 In C/m2? In electrons/m^2? We now consider how some of the macroscopic quantities relate to atomic quantities. Charge is specified as coulombs (C), currents are specified in Amperes (A), Potential differences are specified in Volts. 1 volt=1 J/C. 1W=1J/s 1C=1F/V Remember that each electron carries a charge of -1.6 Times 10 ^-19 coulomb. Almost all the important relationships involve just 3 symbols

Explanation / Answer

2)

charge q1=5uC at y=20cm

charge q2=-5 uC at y=-20cm

a)

at x=0, y=0, potential V=0

b)

workdone W=k*q1*q2/(r)

W=9*10^9*(5*10^-6)^2/(40*10^-2)

W=0.5625 J

W=562.5*10^-3 J

work done is +ve

3)

area of plate, A=1cm^2 =1*10^-4 m^2

separation d=0.02cm =0.02*10^-2 m

battery potential v=90 V

a)

electric field E=v/d

E=90/(0.02*10^-2)

E=4.5*10^5 v/m

b)

if dielectric constant, K=6

but, electric field is remains same, E=4.5*10^5 V/c

c)

capacitance C'=k*C

C'=k*(eo*A/d)

C'=6*8.85*10^-12*(1*10^-2)/(0.02*10^-2)

C'=2.655*10^-9 F

C'=2.655 nF

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