Question

A capacitor of capacitance 'C' is
connected with a cell of 'E' e.m.f. When
it is fully charged, a dielectric medium of
dielectric constant 'k' is introduced in it
find it's final charge if cell is kept
connected with it :-
CE
CE
k k
КСЕ
с
E​

Answer 1

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Answer 2

Given:

A capacitor of capacitance 'C' is

connected with a cell of 'E' e.m.f. When

it is fully charged, a dielectric medium of

dielectric constant 'k' is introduced in it.

To find:

The final charge when the cell is kept connected.

Calculation:

The most important point to remember in this type of questions is that:

• If the cell is kept attached to the capacitor during insertion of the dielectric slab, the potential remains constant.

So, new capacitance be kC; let final charge be q:

$\sf{ \therefore \: V = constant}$

$\sf{ = > \: CQ = (kC)q}$

$\sf{ = > \: C(CE) = (kC)q}$

$\sf{ = > \:CE = kq}$

$\sf{ = > \:q = \dfrac{CE}{k} }$

So, final answer is:

$\boxed{ \bf{\:q = \dfrac{CE}{k} }}$