If I charge a capacitor to 12 V. I disconnect it from the battery, then carefully remove its dielectric material (ε=2ε0)
•What Happens? and why?
please explain in details
Once you disconnect it from battery the charge on capacitor
cannot change There is no path for it to flow so charge will remain
same now
C=KE0A/d
if you remove dielectric then its Capacitance is reduced as K =1
for vacum or air and for dielectric K>>1
V=Q/C,
Q is same and C is decreased so V increases
Electric field=V/d
since, d is same and V is increased so electric field is
increased.
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