a parallel plate capacitor has rectangular plates of 400 CM square and are separated by a distance of 2 mm with air as medium what charge will appear on the plates if a 200 volt potential difference is applied across the condenser

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Solution

DEar Student,

$A = 0.04 m^{2}, d = 0.002 m, V = 200 v C = \frac{ε_{0} A}{d} Q = C V = \frac{ε_{0} A}{d} V = \frac{8.854 * 10^{- 12} * 0.04 * 200}{0.002} = 35.416 * 10^{- 9} C R e g a r d s$

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a parallel plate capacitor has rectangular plates of 400 CM square and are separated by a distance of 2 mm with air as medium what charge will appear on the plates if a 200 volt potential difference is applied across the condenser

DEar Student, A=0.04 m2, d=0.002m, V=200 v C=ε0AdQ=CV=ε0AdV=8.854*10-12*0.04*2000.002=35.416*10-9 CRegards

DEar Student,

$A = 0.04 m^{2}, d = 0.002 m, V = 200 v C = \frac{ε_{0} A}{d} Q = C V = \frac{ε_{0} A}{d} V = \frac{8.854 * 10^{- 12} * 0.04 * 200}{0.002} = 35.416 * 10^{- 9} C R e g a r d s$