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(a) Distinguish, with the help of a suitable diagram, the difference in the behaviour of a conductor and a dielectric placed in an external electric field. How does polarised dielectric modify the original external field?
(b) A capacitor of capacitance C is charged fully by connecting it to a battery of emf E. It is then disconnected from the battery. If the separation between the plates of the capacitor is now doubled, how will the following change?
(i) charge stored by the capacitor.
(ii) field strength between the plates.
(iii) energy stored by the capacitor.
Justify your answer in each case.

 

 

 

 
 
 
 
 

Answers (1)

E_{net}=E_{o}-E_{in}

For conductor E_{o}-E_{in}=0

For Insulator E_{o}-E_{in}\neq 0

Induced electric field due to polarization is opposite in direction to the applied electric field.

b) (i) Charge stored by the capacitor remains the same since after disconnecting the capacitor no charge transfer takes.

(ii) Field strength =\frac{\sigma }{\varepsilon _{o}}=\frac{q}{\varepsilon _{o}A} remains the same as q is the same.

(iii) Energy stored by the capacitor

Energy stored =\frac{q^{2}}{2c}

                        =\frac{q^{2}d}{2\varepsilon _{o}A}

When separation is doubled  energy also doubles.

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Safeer PP

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