Question

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.)

Answer 1

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