Please solve this exam Parallel plate capacitor A=.01 m^2 change 8*10^-6 c using
ID: 2254471 • Letter: P
Question
Please solve this exam
Parallel plate capacitor A=.01 m^2 change 8*10^-6 c using Gauss's law show the electric field is If the separation plate is .2 mm determine optional difference between plate What is the capacitance of capacitor What is the energy stored in the electric field A resistor of 300oms is connected at t=0 how long does it to lose 1/2 of initial charge? An object is placed 4cm in front of convergent lens with a focal length of magnitude 8cm. A concave (convergent) mirror is placed 6 cm behind the lens and has a radius of coverture of 12 cm. Find the location of all images formed and described them (orientation, magnification, real or virtual). Draw a sketch of principle rays required for the first image, only this sketch will be graded so make it clear. Consider a randomly polarized beam at 450 nm and intensity of 100 unites incident on a plane Polaroid sheet whose polarization angle is along the x-axis. A second sheet of Polaroid with its polarization angle at 60 degrees with respect to the x-axis is inserted after the first Polaroid. What is the intensity after both Polaroid sheets? Two spectral lines in a mixture of hydrogen and deuterium gas have wavelength of 656.30 nm and 656.48 nm, respectively, (a) If a grating of 770 ruling mm is used what angle is the spectrum diffracted at in second order? (b) What is the width of the grating to minimally resolve the spectra in second order? A cylindrical region of radius R contains a uniform magnetic field to it axis with magnitude which increase linearly with time. Find the electric field induced at point {a} distance {r} from the cylindrical axis inside the cylinder. Use Ampere's law to find and describe completely the magnetic field distance r from the middle of a long straight wire carrying current I? Show all argument. The diagram below show a light ray traveling from air into material X , what is the index of refraction of X ? (Diagram)Explanation / Answer
a.EF =E = Q/eoA
E = 8*10^-6 /(8.85*10^-12 * 0.11)
E = 9*10^7 N/C
b. V = Ed = 9*10^7 * 0.2*10^-3 = 18000 Volts
c. C = eoA/d = 8.85*10^-12 * 0.01/0.2*10^-3 = 0.44 nF
d. energy U = 0.5 QV = 0.5 * 8*10^-6 * 18000 = 0.072 Joules
e. Q = q0(1-e^t/RC)
1-e^-t/RC = 0.5
e^-t/RC = 0.5
-t/RC = -0.693
t = 0.693 * 300* 0.44*10^-9
t = 91 ns
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