A fully charged parallel-plate capacitor remains connected to a battery while a

Question

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

Explanation / Answer

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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.