Special case: Find the electric potential at P on the y axis for d >> L. What do

Question

Special case: Find the electric potential at P on the y axis for d >> L. What do you expert your result to be?(5 Points.) Replace d by y in your result from part (a) and find an expression for the y-component of the electric field.(5 Points.) A parallel-plate capacitor in air (assume dry air) has plate area of 25cm2 and a plate separation of 2mm. The capacitor is charged to a 300 V and disconnected from the source. A dielectric with k = 4.7 is inserted between the plates, completely filling the volume between the plates. Determine the charge on the capacitor, (5 Points.) the electric field between the plates before the dielectric has been inserted, (3 Points.) the capacitance and potential difference of the capacitor after the- dielectric has been inserted, and (4 Points.) the change in energy of the capacitor. (4 Points.) In Figure 3, C1 = 8 mu F, C2 = C3 = 50 mu F, C1 = 15 mu F, and C3 10 mu F. The- applied potential is Vab = 12 V. Determine the equivalent capacitance of the network between points a and b.and (5 Points.) the charge and potential difference across each capacitor. (10 Points.) An electrical conductor designed to carry large currents has a circular cross section 2.5 mm in diameter and is 14.0 m long. The resistance between its ends is 0.101 Ohm . What is the resistivity of the interval?(4 Points.)

Explanation / Answer

Given that Area of plate A= 25 x 10^-4 m^2 plate seperation is d = 2.0x 10^-3 m potential difference is V= 300 V ------------------------------------------------------- The capacitance of the parallel plate capacitor is              C =Ao/d                  = (25x 10^-4 m^2)(8.85x 10^-12 C2/N.m2)/( 2.0x 10^-3 m)                  = 1.11 x10^-11 F                  = 0.111 pF                  = 0.111 pF a) The charge on the capacitor is        q =CV           = ( 1.11 x10^-11 F)(300 V)           =  33.2 x10^-9 C           = 33.2 nC b) The electric field between the plates before the dielectric has been inserted is         E =V/(kd)            = (300 V)/((4.7)(2.0x 10^-3 m))            = 3.19 x10^4 N/C c) The capacitance of the parallel plate capacitor after the dielectric has been inserted is                 C' =C/k                     = (0.111 pF)/(4.7)                     = 0.0236 pF Potential diffence is              V' =V/k                  = (300 V)/(4.7)                  =63.83 V d) The energy stored in the capacitor is               Uo = (1/2)CV^2                    = (1/2)(1.11 x10^-11 F)(300 V)^2                    = 4.995 x10^-7 J     The energy stored in the capacitor with dielectric is              U =Uo/k                  = (4.995 x10^-7 J)/(4.7)                  = 1.062765  x10^-7 J The change in energy of the capacitor is            U = U- Uo                  = (4.995 x10^-7 J) - (1.062765  x10^-7 J )                  = 3.932x10^-7 J Note:Please submit one question per post.                      Potential diffence is              V' =V/k                  = (300 V)/(4.7)                  =63.83 V d) The energy stored in the capacitor is               Uo = (1/2)CV^2                    = (1/2)(1.11 x10^-11 F)(300 V)^2                    = 4.995 x10^-7 J     The energy stored in the capacitor with dielectric is              U =Uo/k                  = (4.995 x10^-7 J)/(4.7)                  = 1.062765  x10^-7 J The change in energy of the capacitor is            U = U- Uo                  = (4.995 x10^-7 J) - (1.062765  x10^-7 J )                  = 3.932x10^-7 J Note:Please submit one question per post. Note:Please submit one question per post.                     

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