Q.
Five identical plates each of area $A$ are joined as shown in the figure. The distance between the plates is $d.$ The plates are connected to a potential difference of $V\ volts.$ The charge on plates 1 and 4 will be
A) $\frac{\varepsilon_{0}AV}{d},\frac{2\varepsilon_{0}AV}{d}$
B) $\frac{{- \varepsilon}_{0}AV}{d},\frac{2\varepsilon_{0}AV}{d}$
C) $\frac{\varepsilon_{0}AV}{d},\frac{- 2\varepsilon_{0}AV}{d}$
D) $\frac{{- \varepsilon}_{0}AV}{d},\frac{- 2\varepsilon_{0}AV}{d}$
Explanation
(c) The given circuit can be redrawn as follows. All capacitors are identical and each having capacitance $C = \frac{\varepsilon_{0}A}{d}$
[image:1_image401.emf]
|Charge on each capacitor| = |Charge on each plate|
$= \frac{\varepsilon_{0}A}{d}V$
Plate 1 is connected with positive terminal of battery so charge on plate 4 will be $+ \frac{\varepsilon_{0}A}{d}.V$
Plate 4 comes twice and it is connected with negative terminal of battery, so charge on plate 4 will be $- \frac{{2\varepsilon}_{0}A}{d}V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.