Question

Two 3.00 cm × 3.00 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC . Part A What is the electric field strength inside the capacitor if the spacing between the plates is 1.30 mm ?

Answer 1

`8.89\cdot 10^4 V/m`

Step-by-step explanation:

The electric field strength between the plates of a parallel plate capacitor is given by

`E=(\sigma)/(\epsilon_0)`

where

`\sigma` is the surface charge density

`\epsilon_0` is the vacuum permittivity

Here we have

`A=3.00 \cdot 3.00 = 9.00 cm^2 = 9.0\cdot 10^(-4) m^2` is the area of the plates

`Q=0.708 nC = 0.708 \cdot 10^(-9) C` is the charge on each plate

So the surface charge density is

`\sigma=(Q)/(A)=(0.708\cdot 10^(-9))/(9.0\cdot 10^(-4) m^2)=7.87\cdot 10^(-7) C/m^2`

And now we can find the electric field strength

`E=(\sigma)/(\epsilon_0)=(7.87\cdot 10^(-7))/(8.85\cdot 10^(-12))=8.89\cdot 10^4 V/m`