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It is a standard problem to consider a dielectric or a conductor between the parallel plates of a capacitor. But what happens to capacity, voltage, charge, inserting between the plates of an ideal capacitor a charged dielectric or a charged conductor (without contact with the plates)?

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  • $\begingroup$ @HenokhLugo you would get two capacitors in series - almost worthy of a separate question, but It may be covered here $\endgroup$ – tom Dec 12 '14 at 19:59
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If you insert a conductor without touching either plates you end up with two capacitors in series with the widths less than that of the original capacitor you had before you inserted the conductor.

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  • $\begingroup$ and if the conductor is previously charged? $\endgroup$ – scrx2 Dec 13 '14 at 14:10
  • $\begingroup$ Then it depends on whether the terminals are connected to the battery or not. If they are not then the two plates will aquire a charge opposte to that of the conductor inserted. However if the terminals are connected to the battery then the charge on the conductor will be forced to redistribute. The positive charge acumilating at the end near negative terminal thus making the other end of conductor depleted of positive charge etc. $\endgroup$ – SAKhan Dec 13 '14 at 19:40
  • $\begingroup$ Capacitance would not change and you could calculate current as before. $\endgroup$ – SAKhan Dec 13 '14 at 22:57
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There are two case it depends on the battery is connected with constant voltage then There are two main cases:-

1)When Battery is connected $V$=constant

   a)Capacity:-

$$C_{0}=\frac{Q}{V_{0}}$$.For a medium permitivity constant will be written as In general,

$$\epsilon_{m}=\epsilon_{0}K$$($\epsilon_{m}$ the permitivity of medium) So, Potential can be written as $$V_{0}=\frac{q}{4\pi\epsilon_{m}r}$$ I just defined the $\epsilon_{m}$ so $$V=\frac{q}{4\pi\epsilon_{0}Kr}$$ $$V=\frac{V_{0}}{K}$$ So $$C=\frac{Q}{\frac{V_{0}}{K}}$$ So $$C=C_{0}K$$. Hence here

$$C =KC_{0}$$

(Here $C_{0}$ is defined for the air )

b)Charge:- $Q$ is directly propotional to $C$ Hence $$Q=Q_{0}K$$

2) When Battery is not connected Q=constant

a) Potential Again $$C=KC_{0}$$ So V is inversely proportional to the the capacitance so, $$V=\frac{V_{0}}{K}$$ This is the case when the dielectric is added.when it will be remove the cases will change.

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