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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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