Physics · Electrostatic Potential And Capacitance · NEET
The CHARGE is the same on every capacitor in series; the VOLTAGE is different. Here is why: the battery only touches the two outer plates. The inner plates are connected to each other and were neutral to start. When charge +Q appears on the left plate of C1, it pulls -Q onto its right plate. That -Q is taken from the plate of the next capacitor, leaving it +Q, and so on down the chain. Charge is only moved around inside the isolated middle section, so every capacitor ends up with the same magnitude Q. The voltages differ because V = Q/C, and a smaller C with the same Q means a larger V across it.
Because you add the RECIPROCALS: 1/C = 1/C1 + 1/C2. Adding more positive 1/C terms makes 1/C bigger, so C itself gets smaller. Physical reason: putting capacitors in series is like increasing the total plate separation. Capacitance C = ε0A/d falls when d rises, so the combination stores less charge per volt. Quick check: 3 µF and 6 µF in series give (3*6)/(3+6) = 2 µF, which is less than 3 µF.
This confuses many NEET students. For resistors, series adds directly (R = R1 + R2) because resistors block current. For capacitors, the roles flip: capacitance measures how easily charge is stored, which behaves more like conductance (1/R). So capacitors in series use 1/C = 1/C1 + 1/C2 (looks like resistors in parallel), and capacitors in parallel use C = C1 + C2 (looks like resistors in series). Rule to remember: capacitors are the mirror image of resistors.
Step 1: find Ceq from 1/C = 1/C1 + 1/C2 + ... Step 2: the total charge is Q = Ceq * V (V = battery voltage). Step 3: this SAME Q sits on every capacitor. Step 4: the voltage across each one is V_i = Q/C_i. The biggest voltage sits across the SMALLEST capacitor, which is a common exam twist.
Yes. For exactly two capacitors, C = (C1 * C2)/(C1 + C2), read as 'product over sum'. It comes straight from 1/C = 1/C1 + 1/C2 = (C1 + C2)/(C1*C2). Warning: product-over-sum works for TWO capacitors only. For three or more you must add all the reciprocals first.
A 3 µF and a 6 µF capacitor are connected in series across a battery. The equivalent capacitance of the combination is:
Three identical capacitors P, Q and S, each of capacitance C, are connected to a battery of voltage V. P and Q are in series with each other, and this pair is in parallel with S (S directly across the battery). If UP is the energy stored in P and UT is the total energy of the system, the ratio UP/UT is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
1/C = 1/C1 + 1/C2 + 1/C3 + ..., where C is the equivalent (effective) capacitance. For only two capacitors it simplifies to C = (C1*C2)/(C1+C2).
Yes. Every capacitor in a series chain carries the same magnitude of charge Q, because charge is only shifted around the isolated inner plates. The voltage, not the charge, is what divides up.
Voltage divides inversely with capacitance: V_i = Q/C_i. The smallest capacitor gets the largest share of voltage, and the total of all the voltages equals the battery voltage.
Capacitance acts like electrical conductance (1/R), so its rules are the mirror image of resistance. Capacitors in series behave like resistors in parallel, and capacitors in parallel behave like resistors in series.
No. Series capacitance is always smaller than the smallest capacitor in the group. If you calculate a value larger than any single capacitor, you have used the parallel rule by mistake.
