Q.
Consider the arrangement of three metal plates A, B, and C of equal surface area and separation d as shown in figure. The energy stored in the arrangement, when the plates are fully charged, is
A) $\frac{\varepsilon_{0}AV^{2}}{2d}$
B) $\frac{\varepsilon_{0}AV^{2}}{d}$
C) $\frac{{2\varepsilon}_{0}AV^{2}}{d}$
D) $\frac{{3\varepsilon}_{0}AV^{2}}{2d}$
Explanation
(b) The arrangement behaves as a combination of 2 capacitors each of capacitance $C = \frac{\varepsilon_{0}A}{d}.$
Thus, equivalent capacity$\ = 2C$
$\therefore\ \ $total energy stored $U = \frac{1}{2} \times (2C)V^{2} = CV^{2} = \frac{\varepsilon_{0}A}{d}V^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.