Question

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?

Answer 1

Answer:

U= $0.112 x10^{-6}$ J

u= 0.708 $\frac{J}{m^{3} }$

Explanation:

diameter= 2 cm

$r= 1 cm * \frac{1m}{100cm} = 0.01 m$

distance= 0.5 mm

$d= 0.5 mm * \frac{1m}{1000m}= 0.5 x10^{-3}$

Area A= $\pi *r^{2}= 0.314x10^{-3} m^{2}$

Volume v= $A*d = 0.314 x10^{-3}m^{2}*0.5x10^{-3} m =0.157x10^{-6} m^{3}$

$v= 0.157 x10^{-6} m^{3}$

Constant vacuum permittivity

$E_{o}= 8.85x10^{-12}$

a).

$C= \frac{A*E_{o} }{d} = \frac{0.314x10^{-3} *8.85x10^{-12} }{0.5x10^{-3} } \\C= 5.56 x10^{-12}$F

$U= \frac{1}{2} *C *V^{2}\\ U=\frac{1}{2} * 5.56x10^{-12}*(200)^{2} \\U=0.112 x^{-6} J$

b).

$u=\frac{U}{v}$

$u=\frac{0.112 x10^{-6}J}{0.157x10^{-6}m^{3} } \\u=0.708 \frac{J}{m^{3} }$