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LekaFEV [45]
3 years ago
9

The plates of a parallel-plate capacitor each have an area of 0.1 m^2 and are separated by a 0.9 mm thick layer of porcelain. Th

e capacitor is connected to a 14 V battery. (The dielectric constant for porcelain is 7.) (a) Find the capacitance. (b) Find the charge stored. (c) Find the electric field between the plates.
Physics
1 answer:
podryga [215]3 years ago
4 0

Answer: a) 6.88 nF; b) 96 .32 nC; c) 108.83 * 10^3 N/C

Explanation: In order to solve this problem we need to used the expresion for the capacitor of two parallel plates, which is given by:

C=εo*A/d;  A is the area and d the separaction between the plates.

inside the capacitor there is a porcelain layer so we have to multiply by the dielectric constant (k=7).

Then C=8.85*10^-12*7*0.1 /0.0009=6.88 nF

Also we know that ΔV=Q/C

Q= C*ΔV= 6.88 nF* 14 V= 96 .32 nC

Finally, The electric field between the plates is given by:

E= σ/εo= Q/(A*εo)= 96.32 /(0.1*8,85* 10^-12)= 108.83 * 10^3 N/C

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0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.

<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>

Position of n the order fringe = n λ D / d

for n = 2

position = 2 λ D / d

λ = 710 nm , D = 1.5m

d = .65 x 10⁻³

position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³

= 3276.92 x 10⁻⁶ m

= 3.276x 10⁻³ m

= 3.276mm .

For λ = 660 nm

position = 2 λ D / d

λ = 660 nm , D = 1.5 m

d = .65 x 10⁻³

position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³

= 3046.15 x 10⁻⁶ m

= 3.046 x 10⁻³ m

= 3.046 mm .

Difference between their position

= 3.276mm ₋ 3.046 mm

= 0.23 mm .

To know more about Fringes refer to:  brainly.com/question/15649748

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