Show that for the capacitor in parallel as shown in figure the capacitance is $C_{eq}=\frac{\epsilon_{0}\epsilon_{r1}A_{1}}{d}+\frac{\epsilon_{0}\epsilon_{r2}A_{2}}{d}=C_{1}+C

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Show that for the capacitor in parallel as shown in figure the capacitance is

$C_{eq}=\frac{\epsilon_{0}\epsilon_{r1}A_{1}}{d}+\frac{\epsilon_{0}\epsilon_{r2}A_{2}}{d}=C_{1}+C_{2}$

written 7.1 years ago by

govindkedar • 280

modified 3.3 years ago by

sagarkolekar ★ 11k

electromagnetic field and wave theory

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