Capacitor – JEE Main Previous Year Questions with Solutions
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Therefore K =2 
(1) 100 sec and 150 sec (2) 150 sec and 200 sec (3) 0 and 50 sec (4) 50 sec and 100 sec [AIEEE-2012] 
[JEE-Mains-2015] 
$\mathrm{C}_{\mathrm{eq}}=\frac{3 \mathrm{C}}{3+\mathrm{C}} \quad \mathrm{Q}=\mathrm{Ceq} \mathrm{E}$ $\mathrm{Q}_{2}=\frac{2 \mathrm{CE}}{(\mathrm{C}+3)}=\frac{3 \mathrm{CE}}{(\mathrm{C}+3)}$ $=2 \mathrm{E}\left[1-\frac{3}{\mathrm{C}+3}\right]$ $\mathrm{Q}_{2 / \mathrm{when}} \mathrm{C}=1_{\mathrm{HF}}=2 \mathrm{E}\left[\frac{1}{4}\right]=\frac{\mathrm{E}}{2}$ $\mathrm{Q}_{2 / \mathrm{when}} \mathrm{C}=3_{1 \mathrm{F}}=2 \mathrm{E}\left[\frac{1}{2}\right]=\mathrm{E}$ $\frac{\mathrm{dQ}}{\mathrm{d} \mathrm{C}}=-\frac{2 \mathrm{E} \cdot 3}{(\mathrm{C}+3)}(-1)=\frac{6 \mathrm{E}}{(\mathrm{C}+3)^{2}}>0$ $\frac{\mathrm{d}^{2} \theta}{\mathrm{d} \mathrm{C}^{2}}=\frac{6 \mathrm{E}(-2)}{(\mathrm{C}+3)^{3}}=-\frac{12 \mathrm{E}}{(\mathrm{C}+3)^{2}}<0$ 
(1) 480 N/C (2) 240 N/C (3) 360 N/C (4) 420 N/C [JEE-Mains-2016] 
$\mathrm{Q}=24+18=42 \mu \mathrm{c}$ $\mathrm{E}=\frac{\mathrm{KQ}}{\mathrm{r}^{2}}$ $\Rightarrow \mathrm{E}=\frac{9 \times 10^{9} \times 42 \times 10^{-6}}{(30)^{2}}=420 \mathrm{N} / \mathrm{C}$ 
$\Rightarrow$ Minimum no. of capacitors $=8 \times 4=32$ 
$\mathrm{Q}=(\mathrm{k} \mathrm{C}) \mathrm{V}$ $=\left(\frac{5}{3} \times 90 \mathrm{pF}\right)(20 \mathrm{V})$ = 3000 pC = 3nC induced charges on dielectric $\mathrm{Q}_{\mathrm{ind}}=\mathrm{Q}\left(1-\frac{1}{\mathrm{K}}\right)=3 \mathrm{nC}\left(1-\frac{3}{5}\right)=1.2 \mathrm{nC}$
Frequently Asked Questions
Find answers to common questions.
What happens to capacitance when a dielectric is inserted?
When a dielectric material with dielectric constant K is inserted between the plates of a capacitor, the capacitance increases by a factor of K: C_new = KC₀. If the capacitor is connected to a battery, the charge stored also increases by K. If it is isolated, the voltage across the plates decreases by a factor of K.
What is the formula for charge on a capacitor during charging?
The charge on a capacitor during charging through a resistor R is Q = Q₀(1 – e^{–t/RC}), where Q₀ = CV is the final (maximum) charge, t is time in seconds, R is resistance in ohms, and C is capacitance in farads. At t = RC (one time constant), the capacitor is approximately 63% charged.
How many questions come from Capacitors in JEE Main every year?
JEE Main typically has 1 to 2 questions from the Capacitors chapter every year. Based on NTA's pattern from 2010 to 2024, RC circuits and parallel plate capacitors with dielectrics are the two sub-topics most likely to appear. Mastering these sub-topics alone can reliably secure 4–8 marks in the Physics section.
Is the Capacitors chapter important for JEE Advanced as well?
Yes. Capacitors appear in JEE Advanced with higher complexity — often combined with electric fields, conductors, and multi-loop circuits. Concepts like capacitor with a conducting slab, force between plates, and energy methods are frequently tested. Solving JEE Main PYQs first builds the foundation; then move to JEE Advanced problems from 2015 onwards for deeper practice.
What is the minimum number of capacitors needed for high-voltage circuits?
To handle a voltage V_total using capacitors each rated at V_max, you need at least ⌈V_total / V_max⌉ capacitors in series per string. The number of parallel strings depends on the required total capacitance. Always round up — partial capacitors are not allowed — and verify that the total capacitance requirement is also met.
How do you find the time constant from a V-t discharge graph?
The time constant τ is the time at which the voltage falls to 37% (0.37) of its initial value V₀. On the graph, find V = 0.37 × V₀, then read off the corresponding time on the x-axis. That time value is τ. This graphical method was directly tested in the AIEEE 2012 question on this page