In: Physics
A fully charged parallel-plate capacitor remains connected to a battery while a dielectric is slid between the plates. Do the following quantities increase, decrease, or stay the same?
(a) C
---Select--- increases decreases stays the same
(b) Q
---Select--- increases decreases stays the same
(c) E between the plates
---Select--- increases decreases stays the same
(d) ?V
---Select--- increases decreases stays the same
(e) energy stored in the capacitor
---Select--- increases decreases stays the same
a)
c increases .
b)
q increases .
c)
E between the plates remains same
d)
?V remains the same
e)
energy stored in the capacitor increases
Explanation:
when a dielectric is slided it has a certain dielectric constant
which is greater than that of air (Eo) , k is ratio of
dielectric/Eo
thus q=cv
and c can be determined , u can find mathemetical explanation in a
book but theorotically when a dielectric is slided the electrons
that are tightly bound in the material form a dipole in the
substance thus balancing the charge of opposite plate and thus more
charge can now be mounted on plate, thus q increases and c
increases .
v remains same because it depends on distance of the plates and
charge the effective distance is lowered and charge is increased
thus no change is observed ( v= k x q/r) r=distance also q=cv thus
v=q/c and both increase such that there is no effect on v . e
remains same between plates because additional field is countered
by dielectric