Question

The electrodes of a parallel plate capacitor are squares of side 25 cm, separated by 1.5 mm. The gap is filled with a dielectric material of relative permittivity 6.8. What is the plate charge when the potential difference between the plates is 140 V?

Answer 1

`.351\mu C`

Step-by-step explanation:

The side of the square = 25 cm =0.25 m

Area of square
`A=side^2=.25^2=0.00625m^2`

Separation between the plate d = 1.5 mm
`=1.5* 10^(-3)m`

Relative permitivity k =6.8

The capacitance of the parallel plate capacitor is given by
`C=(k\epsilon _0A)/(d)=(6.8* 8.85* 10^(-12)* 0.00625)/(1.5* 10^(-3))=2.5* 10^(-9)F`

Voltage across the capacitor V =140 Volt

So charge Q = CV
`=2.5* 10^(-9)* 140=351* 10^(-9)C=.351\mu C`