The voltage drop across each capacitor is 134 V.
Absent an external voltage source, a capacitor is nothing more than a neutral conductor (before charging). But a capacitor starts to store electric charges when an external voltage is placed across it. Currently, the electric charge on a capacitor is precisely proportional to the voltage across it. A change in charge applied to a capacitor causes a change in the voltage across it. Therefore, the voltage across a capacitor rises as it charges. The voltage across the capacitor becomes constant once the capacitor has fully charged. The capacitor will now start to discharge if we remove the external battery. The capacitor's voltage across it drops as it is being discharged, and after a while, it reaches zero.
Given,
C = 4.00pF
C₁ = 8.00pF
V = 400 V
C(eq) = 1/ (114 + 118)
= 2.66667 pF
Q = C * V
Q = 2.66667 * 10⁻¹² F * 400 V
Q₁ = 1.07 nC
Q₂ = 1.07 nC
V = Q/C = 1.07 * 10⁻⁹ C/ 4.00 * 10⁻¹² F
V₁ = 267.0 V
V₂= Q/C
= 1.07 * 10⁻⁹ C / 8.00 * 10⁻¹² F
V₂ = 133.0 V
Thus, the voltage drop is, V₁ - V₂ = 267.0 - 133.0
Voltage drop = 134 V
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