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 betw…

Question

asked Mar 3, 2020 16.5k views

1 Answer

Arrehman · Answer 1 · 2020-03-09T20:59:12+0000

Answer:

$8.89\cdot 10^4 V/m$

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

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