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Irina-Kira [14]
2 years ago
12

Two parallel plates 19 cm on a side are given equal and opposite charges of magnitude 2.0 ✕ 10^−9 C. The plates are 1.8 mm apart

. What is the electric field (in N/C) at the center of the region between the plates? (Enter the magnitude.)
Physics
1 answer:
Romashka [77]2 years ago
5 0

Answer:

E   = 5291.00 N/C

Explanation:

Expression for capacitance is

C = \frac{\epsilon  A}{d}

where

A is area of square plate

D = DISTANCE BETWEEN THE PLATE

C = \frac{\epsilon\times(19\times 10^{-2})^2}{1.5\times 10^{-3}}

C = 24.06 \epsilon

C = 24.06\times 8.854\times 10^{-12} F

C =2.1\times 10^{-10} F

We know that capacitrnce and charge is related as

V = \frac{Q}{C}

 = \frac{2\tiimes 10^{-9}}{2.\times 10^{-10}}

v = 9.523 V

Electric field is given as

E = \frac{V}{d}

   = \frac{9.52}{1.8*10^{-3}}

E   = 5291.00 V/m

E   = 5291.00 N/C

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Nuclear power plants provide a stable base load of energy. This can work synergistic with renewable energy sources such as wind and solar. The electricity production from the plants can be lowered when good wind and solar resources are available and cranked up when the demand is high.

 

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