# Electrical Force between plates(capacitor)

by haleycomet2
Tags: capacitor, force
 P: 29 we know that electrical field strength,E,can be counted by equation 1. E=V/d,or 2. E=Qq/4$$\pi$$$$\epsilon$$r 2 In the circuit with capacitor, the potential difference ,V is decided by battery.So from the first equation ,the E is not affected by permittivity(dielectric material) but it does affected according to second equation. What is the correct explanation of the electrical force between plates??
 P: 29 Electrical Force between plates(capacitor) Thanks a lot for your reply. I went to check the derivation of the formula E=V/d.Referring to the graph,the work done to move charge A to B can be expressed by: W=qEd or W=qVAB Therefore by eliminating q,VAB=Ed,so E=VAB/d Besides,i also think other derivation,in the circuit consist of capacitor,using the formulas below: E=$$\sigma$$ /$$\epsilon$$ C=$$\epsilon$$ A/d Q=CV $$\sigma$$ =Q/A E=$$\sigma$$ /$$\epsilon$$ =(Q/A)/$$\epsilon$$ =(CV/A)/$$\epsilon$$ =($$\epsilon$$ AV/Ad)/$$\epsilon$$ =V/d Derivation above showed that the electric field strength is independent of $$\epsilon$$,dielectric material!?If according to the field strength formula E=kq/r2,it seems that the field strength is independent of capacitor area?! ps:The charge Q counted from Q=CV is the amount of charge on one side of plate or both side of the capacitor? Thank you again...=) Attached Thumbnails