Question

A capacitor is constructed from two square, metallic plates of sides land separation d. Charges +Qand-Q are placed on the plates, and the power supply is then removed. A material of dielectric constant x is inserted a distance x into the capacitor as shown in the figure below. Assume d is much smaller than x. Suggestion: The system can be considered as two capacitors connected in parallel. (Use the following as necessary: 80, X, 1, Q. d, and x.) + + + + + + - - - O (a) Find the equivalent capacitance of the device. - 9 ( +8 (3)(x-1)) (b) Calculate the energy stored in the capacitor. g².a 2.El+fix:(k-1)) (c) Find the direction and magnitude of the force exerted by the plates on the dielectric. 02.0-6.(x-1) magnitude 2.60(+.x=(x-1))? direction + D (d) Obtain a numerical value for the force when x = 1/2, assuming l = 5.10 cm, d = 1.90 mm, the dielectric is glass (K = 4.50), and the capacitor was charged to 1.90 x 10' V before the dielectric was inserted. magnitude 5.48 direction Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. WN + D

Answer 1

We have force, E= Q² dl (K-1) 260 (ľ + d. x.(K-1))" :F= t 14 v2 (dl(1-1)) (d) (2 E) ( 1²7 l.x (k-1)) F = to l V² (6-4) ad(e+l.a.(k-1)) -12 given l = 5,1cm = 0.051m. a= = 0.0255 m, d = 1,98103 m = 0,0019 m K = 4.5. V= 1900 V., E= 8,85 x10th F = 8,850102 x (5.1X102) 5 x 11,9x103) ² (415-1). 2 x 19x10 [10.050) +10.051X0.0255)(3.5))* = 38806.373 x 1016 194.41 x 10 E - 1984.498% 10N = 198 x 10 GN = F = 198 MN = 198 x 100 N = 0.000198 N /