Capacitor | Question Bank for Class 12 Physics
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Get important questions of Capacitor for Boards exams. Download or View the Physics Question Bank Class 12. These important questions will play a significant role in clearing concepts of Physics. This question bank is designed by NCERT, and the questions are updated with respect to the upcoming Board exams. You will get all the important questions for the Class 12 Physics chapter-wise CBSE. Click Here for Detailed Chapter-wise Notes of PHYSICS for Class 12th, JEE & NEET. ‘q’ with time ‘t’ when a condenser charged. India's Best Exam Preparation for Class 12th - Download Now
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Also two capacitors of $10 \mu F$ are in parallel. Their equivalent capacitance is $C_{2}=10+10=20 \mu F . C_{1}$ and $C_{2}$ are mutually parallel, so their equivalent capacitance $C_{12}=C_{1}+C_{2}=20+20=40 \mu F$ Now $C_{12}$ and $40 \mu F$ are in series, hence equivalent capacitance of network is given by $\frac{1}{C_{A B}}=\frac{1}{C_{12}}+\frac{1}{40}=\frac{1}{40}+\frac{1}{40}=\frac{1}{20} \Rightarrow C_{A B}=20 \mu F$ Charge $Q=C V=20 \times 10^{-6} \times 100=2 \times 10^{-4} C$ 
$\therefore$ Total charge, $Q=Q_{1}+Q_{2} \ldots(i)$ But $Q_{1}=C_{1} V_{1}$ and $Q_{2}=C_{2} V_{2}$ $\therefore$ By equation (i) we get $Q=C_{1} V_{1}+C_{2} V_{2}$ Total capacity $C=C_{1}+C_{2}$ (i) Common potential: Let the conductors are connected by a wire and the common potential becomes $V$. 
(ii) Loss of Energy: Total energy of the conductors before connection 
since is always positive therefore will be always positive. Hence in the distribution of charges there is always a loss of energy. $\therefore$ Loss energy 
$E_{+}=\frac{\sigma}{2 \varepsilon_{0}}=\frac{Q}{2 A \varepsilon_{0}}$ at all points if the plate is large. The negative charge $-Q$ finds itself in the field of this positive charge. The force $-Q$ on is, therefore $\cdot F=-Q E$$=(-Q) \frac{Q}{2 A \varepsilon_{0}}=-\frac{Q^{2}}{2 A \varepsilon_{0}}$ The magnitude of the force is is $F=\frac{Q^{2}}{2 A \varepsilon_{0}}$ (ii) Work done in increasing the separation between the plates by a distance $d x$ against the force of attraction $F$ between the plates is $d W=F d x \cdots$ (i) As, energy density, $u=\frac{1}{2} \varepsilon_{0} E^{2}$ $\therefore$ Increase in potential energy m $=u A d x=\frac{1}{2} \varepsilon_{0} E^{2} A d x \ldots$ (ii) From the equation (i) and (ii) 
$F d x=\frac{1}{2} \varepsilon_{0} E^{2} A d x \Rightarrow F=\frac{1}{2} \varepsilon_{0} E^{2} A=\frac{1}{2}\left(\varepsilon_{0} A E\right) E$ $\Rightarrow F=\frac{1}{2}\left(\varepsilon_{0} A \frac{V}{d}\right) E \Rightarrow F=\frac{1}{2}(C V) E \Rightarrow F=\frac{1}{2} Q E$ The factor $\frac{1}{2}$ is in the expression due to the fact that plates are being moved against an average value $\frac{E}{2} .$ 
As in case of conducting slab $E_{i}=E_{0} .$ (Net electricfield inside the conducting slab is zero $) .$ Now potential, difference between the plates of capacitor is $V=E_{0}(d-t)=\frac{\sigma}{\varepsilon_{0}}(d-t)$$\Rightarrow Q=\sigma A \Rightarrow C=\frac{Q}{V}=\frac{\varepsilon_{0} A}{d-t}$ 
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Fundamental Formulas at a Glance
| Quantity | Formula | Condition |
|---|---|---|
| Capacitance (parallel plate) | $C = \dfrac{\varepsilon_0 A}{d}$ | Air/vacuum between plates |
| Capacitance with a dielectric | $C = K \cdot \dfrac{\varepsilon_0 A}{d}$ | Dielectric constant $K$ |
| Charge stored | $Q = CV$ | — |
| Energy stored | $U = \dfrac{1}{2}CV^2 = \dfrac{Q^2}{2C}$ | — |
| Series combination | $\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \cdots$ | — |
| Parallel combination | $C_{eq} = C_1 + C_2 + \cdots$ | — |
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Frequently Asked Questions
Find answers to common questions.
What is the difference between series and parallel capacitor combinations?
In a series combination, all capacitors carry the same charge, and the reciprocal of the equivalent capacitance equals the sum of reciprocals: $1/C_{eq} = \Sigma(1/C_i)$ — the equivalent is always less than the smallest individual capacitance. In a parallel combination, all capacitors share the same voltage, and $C_{eq} = \Sigma C_i$ — the equivalent is always greater than the largest individual capacitance.
How many capacitor questions come in CBSE Class 12 Board exams?
Capacitors typically appear in 2–3 questions in CBSE Board exams — usually one 2-mark conceptual question, one 3-mark derivation or effect of dielectric question, and one 5-mark numerical on circuit combinations or energy. The chapter also overlaps with the Electrostatics unit which is a compulsory question set in Section B and C of the paper.
Why does energy decrease when a dielectric is inserted after the battery is disconnected?
When the battery is disconnected, charge Q is fixed. Inserting a dielectric increases capacitance K times, so energy $U = Q^2/2C$ decreases by a factor of K. The "lost" energy is not destroyed — it is converted into mechanical work done as the dielectric slab is pulled into the capacitor by the attractive electric force between the slab's dipoles and the fringing field.
What is the SI unit of capacitance and what does 1 Farad mean?
The SI unit of capacitance is the Farad (F), named after Michael Faraday. 1 Farad means a conductor gains a potential of 1 Volt when 1 Coulomb of charge is placed on it. In practice, most capacitors used in circuits are in the microfarad (μF) or picofarad (pF) range since 1 Farad is an extremely large capacitance.
How do I prepare capacitor questions for JEE Main effectively?
Start with NCERT examples and exercises to build conceptual clarity, then solve NCERT Exemplar problems for higher-order thinking. Focus specifically on the two boundary condition scenarios (battery connected vs. disconnected) and practice circuit simplification with 4–6 capacitors. eSaral's IIT Bombay faculty — including AIR-41 ranked teachers from Kota — offer structured video lectures and Kota-quality practice sets for exactly these patterns.
What is the formula for a parallel plate capacitor with a dielectric slab of thickness t (not filling the full gap d)?
When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted in a gap of $d$ (where $t < d$):
$$
C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}
$$
If $t = d$ (slab fills the entire gap), the formula reduces to:
$$
C = \frac{K\varepsilon_0 A}{d}
$$
This is a common JEE problem variation not always seen in standard textbooks.