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Category > Statistics Posted 02 Jun 2017 My Price 7.00

Two parallel conducting plates, each having a surface area of 2 × 10−2 m2 on its sides

Two parallel conducting plates, each having a surface area of 2 × 10−2 m2 on its sides, form a parallel-plate capacitor. Their separation is 1.25 mm and the medium between them is free space. A 100-V d.c. battery is connected across them, and it is maintained there at all times. Then a dielectric sheet, 1 mm thick and with the same shape and area as the plates, is slipped carefully between the plates so that one of its sides touches one of the conducting plates. After the insertion of the slab and neglecting fringing, if the dielectric constant of the dielectric sheet is εr = 5, determine the:

 

(a) Electric field intensity between the plates (inside and outside the slab).

(b) Electric flux density between the plates (inside and outside the slab).

(c) Surface charge density in each of the plates.

(d) Total charge in each of the plates.

(e) Capacitance across the slab, the free space, and both of them.

(f) Energy stored in the slab, the free space, and both of them.

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Status NEW Posted 02 Jun 2017 10:06 AM My Price 7.00

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file 1496399233-1752916_1_636319130155109991_In-parallel-plate-capacitor-arrangement--two-metal-plates-of-equal-area-are--separated-by-a-distance--d.docx preview (760 words )
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