info@clay6.com
+91-9566306857
(9am to 6pm)
logo

Ask Questions, Get Answers

X
Want help in doing your homework? We will solve it for you. Click to know more.
 
Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Electromagnetic Waves

A parallel plate capacitor with plate area $A$ and separation $d$ between the plates is charged by a constant current $i$. Consider a plane surface of area $A/4$ parallel to the plates and drawn symmetrically between the plates. Calculate the displacement current through this area.

$\begin {array} {1 1} (a)\;i/4 & \quad (b)\;i \\ (c)\;i/2 & \quad (d)\;i/3 \end {array}$

 

1 Answer

Need homework help? Click here.
If $q$ be the charge on the capacitor at time $t_1$, then electric field between plates of the capacitor,
$ E = \large\frac{q}{ \in_oA}$
Flux through the area, $ \phi E = \bigg(\large\frac{q}{\in_oA} \bigg) \bigg( \large\frac{A}{4} \bigg) $$= \large\frac{q}{4\in_o}$
Displacement current, $I_D = \in_o \large\frac{ d\phi E}{dt}$$ = \in_o \large\frac{d\bigg( \large\frac{q}{4}\in_o\bigg)}{dt}$$ = \large\frac{1}{4}\large\frac{ dq}{dt}$$ = \large\frac{i}{4}$
Ans : (a)
answered Feb 27, 2014 by thanvigandhi_1
 

Related questions

...