May I please get help with question12) a and b . Thanks . The separation of the
ID: 3161457 • Letter: M
Question
May I please get help with question12) a and b . Thanks . The separation of the plates of a plate is doubled while the charge on them is held constant. What happens to each of the following? The electric field between the plates The potential difference across the plates. The capacitance of the capacitor. The energy stored in the capacitor. What is a dielectric? Exactly what does the dielectric constant measure? What are its units? A 20.0 pF parallel plate capacitor has a 1. Mm thickness of paper completely filling the gap between the plates. If the dielectric strength of paper is 16.0 times 10^6 V/m what is the gap maximum charge that can be placed on the capacitor?Explanation / Answer
A) The initial conditions are
A= Area of plates (does not change), d0=distance between plates, 0=permittivity of free space,
Q= charge (constant)
Capacitance C0= 0A/d0
Potential V0=Q/C0
Electric field E0= V0/d0
Energy En0= ½ C0 V02
After the plate separation is doubled
d=2d0
c) C= 0A/d= 0A/ ( 2d0)= C0/2 i.e. capacitance has halved
b) V=Q/C=Q/(C0/2)=2.Q/C0=2V0 i.e. potential has doubled
a) E=V/d= 2V0/2d0=E0 i.e. Electric field remains same
d) Energy En=½ CV2= ½.( C0/2 ).( 2V0)2= 2. (½ C0 V02)= ½ En0, i.e. Energy has halved
B) Capacitance C= 20 pF= 20 x 10-12 F
Distance d= 1mm= 10-3 m
The break down field for paper Emax= 16 x 106 V/m
Vmax== Emax/d=16 x 106/10-3 = 16 x 109 V/m
Maximum charge Qmax= Vmax C= 16 x 109 x 20 x 10-12 =320 x 10-3 =320 mC