IIT JEE , AIEEE Forum - Physics , Electricity - bit easy question , but still ...want help.

padio

What is force acting on a dielectric introduced in a capacitor?.............IF capacitor remains connected to a charging source? and if battery is disconnected after the capicitor is charged? PLESE if u could explain a little bit more ...on ur answer .

akumar_ak

if E× be the electric field intensity without the dielectric,and E with it
then the force
charge = q,

F = (E×-E)q

krishna.gopal

Suppose at any instant the portion of dielectric inside capacitor is x and we make it x+dx (ie. we insert dielectric into it). This will increase its capacitance and there will be an change in Potential .Energy (U) and F = -dU/dx

If Capacitor is connected to batery its V is constant and PE increases So dielectric will be pushed out

If Capacitor is disconnected to batery its Q is constant and PE decreases So dielectric will be pulled in

imnotserious_2008

WHEN BATTERY IS REMOVED AFTER CHARGING

if the battery is removed then the charge stored in the capacitor becomes constant
q=constant
suppose a dilectric slab of dielectric constant x is introduced
we know
E=E(not)/k
E is proportional to 1/k (..:E(not) is constant if q is constant)
now E will become 1/x times
f=qE q is constant thus F will become i/x times

imnotserious_2008

WHEN BATTERY REMAINS CONNECTED

IF THE BATTERY REMAINS CONNECTED,THE POTENTIAL DIFFERENCE V BECOMES CONSTANT.

E=V/D
so E will become 1/times the distance between plates is changed

now Q=CV
FIND THE CHANGE IN CAPACITANCE WHICH WILL BE PROPORTIONAL TO K/D
AND FINALLY GET F=QE

PLEASE RATE IF U R SATISFIED

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