A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates.
A sheet having thickness left(frac13right)^mathrmrd$\left(\frac{1}{3}\right)^{\mathrm{rd}}$ of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :
### Related Formula
C = fracepsilon_0 Ad - t + fractK$$C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$$
Alternatively, treat as two capacitors in series: C_eq = fracC_1 C_2C_1 + C_2$C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$
### Core Logic
Initial capacitance (vacuum): C = fracA epsilon_0d$C = \frac{A \epsilon_{0}}{d}$.
Capacitance with Dielectric diagram for Q34 - JEE Main 2026 Morning
The system can be modelled as two capacitors in series:
1. A vacuum capacitor of thickness d - fracd3 = frac2d3$d - \frac{d}{3} = \frac{2d}{3}$
2. A dielectric capacitor of thickness fracd3$\frac{d}{3}$ and dielectric constant K$K$.
Capacitance with Dielectric diagram for Q34 - JEE Main 2026 MorningCapacitance with Dielectric diagram for Q34 - JEE Main 2026 Morning
### Step 1: Capacitors in Series
The capacitances are:
C_1 = fracepsilon_0 Aleft(frac2d3right) = frac32 left(fracepsilon_0 Adright) = frac32 C$$C_1 = \frac{\epsilon_0 A}{\left(\frac{2d}{3}\right)} = \frac{3}{2} \left(\frac{\epsilon_0 A}{d}\right) = \frac{3}{2} C$$C_2 = fracK epsilon_0 Aleft(fracd3right) = 3K left(fracepsilon_0 Adright) = 3KC$$C_2 = \frac{K \epsilon_0 A}{\left(\frac{d}{3}\right)} = 3K \left(\frac{\epsilon_0 A}{d}\right) = 3KC$$
Now, equivalent capacitance in series:
C_texteq = fracC_1 C_2C_1 + C_2 = fracleft(frac32 Cright) times (3KC)frac32 C + 3KC$$C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{\left(\frac{3}{2} C\right) \times (3KC)}{\frac{3}{2} C + 3KC}$$C_texteq = fracfrac92KC^2frac32C(1 + 2K)$$C_{\text{eq}} = \frac{\frac{9}{2}KC^2}{\frac{3}{2}C(1 + 2K)}$$C_texteq = frac3KC2K + 1$$C_{\text{eq}} = \frac{3KC}{2K + 1}$$
### Pattern Recognition
Partial dielectric filling of thickness t$t$: Use formula C_textnew = fracepsilon_0 Ad - t + t/K$C_{\text{new}} = \frac{\epsilon_0 A}{d - t + t/K}$. Plugging t = d/3$t = d/3$ directly gives fracepsilon_0 Ad - d/3 + d/3K = fracepsilon_0 Afrac2d3 + fracd3K = frac3K epsilon_0 A2Kd + d = frac3KC2K+1$\frac{\epsilon_0 A}{d - d/3 + d/3K} = \frac{\epsilon_0 A}{\frac{2d}{3} + \frac{d}{3K}} = \frac{3K \epsilon_0 A}{2Kd + d} = \frac{3KC}{2K+1}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Capacitance with Dielectric diagram for Q34 - JEE Main 2026 Morning
More Electrostatics Previous-Year Questions — Page 3
Q23jee_main_2025_08_april_eveningCapacitance
Space between the plates of a parallel plate capacitor of plate area 4mathrm~cm^2$4\mathrm{~cm}^{2}$ and separation of (d) 1.77mathrm~mm$1.77\mathrm{~mm}$, is filled with uniform dielectric materials with dielectric constants (3$3$ and 5$5$) as shown in figure. Another capacitor of capacitance 7.5mathrm~pF$7.5\mathrm{~pF}$ is connected in parallel with it. The effective capacitance of this combination is ________ mathrm~pF$\mathrm{~pF}$.
This diagram shows a parallel plate capacitor filled with two layers of dielectric constant k=5 and k=3, each having thickness d/2.
(Given varepsilon_0 = 8.85times 10^-12mathrm~F/m$\varepsilon_0 = 8.85\times 10^{-12}\mathrm{~F/m}$)
Numerical Answer.Answer: 15 to 15
Solution
### Related Formula
C = k fracvarepsilon_0 At$$C = k \frac{\varepsilon_0 A}{t}$$frac1C_textseries = frac1C_1 + frac1C_2$$\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2}$$C_textparallel = C_texteq + C_p$$C_{\text{parallel}} = C_{\text{eq}} + C_p$$
### Core Logic
The dielectrics k_1 = 5$k_1 = 5$ and k_2 = 3$k_2 = 3$ split the capacitor separation horizontally into two layers, each of thickness fracd2$\frac{d}{2}$. Thus, they act as two capacitors connected in **series**:
- Capacitor 1: C_1 = k_1 fracvarepsilon_0 Ad/2 = 5 times frac2 varepsilon_0 Ad = 10 fracvarepsilon_0 Ad$C_1 = k_1 \frac{\varepsilon_0 A}{d/2} = 5 \times \frac{2 \varepsilon_0 A}{d} = 10 \frac{\varepsilon_0 A}{d}$
- Capacitor 2: C_2 = k_2 fracvarepsilon_0 Ad/2 = 3 times frac2 varepsilon_0 Ad = 6 fracvarepsilon_0 Ad$C_2 = k_2 \frac{\varepsilon_0 A}{d/2} = 3 \times \frac{2 \varepsilon_0 A}{d} = 6 \frac{\varepsilon_0 A}{d}$
### Step 1: Compute Base Capacitance Factor
First, find the term fracvarepsilon_0 Ad$\frac{\varepsilon_0 A}{d}$ in SI units:
- A = 4mathrm~cm^2 = 4 times 10^-4mathrm~m^2$A = 4\mathrm{~cm}^2 = 4 \times 10^{-4}\mathrm{~m}^2$
- d = 1.77mathrm~mm = 1.77 times 10^-3mathrm~m$d = 1.77\mathrm{~mm} = 1.77 \times 10^{-3}\mathrm{~m}$
- varepsilon_0 = 8.85 times 10^-12mathrm~F/m$\varepsilon_0 = 8.85 \times 10^{-12}\mathrm{~F/m}$fracvarepsilon_0 Ad = frac8.85 times 10^-12 times 4 times 10^-41.77 times 10^-3 = frac35.4 times 10^-161.77 times 10^-3 = 20 times 10^-13mathrm~F = 2mathrm~pF$$\frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 4 \times 10^{-4}}{1.77 \times 10^{-3}} = \frac{35.4 \times 10^{-16}}{1.77 \times 10^{-3}} = 20 \times 10^{-13}\mathrm{~F} = 2\mathrm{~pF}$$
Now find C_1$C_1$ and C_2$C_2$:
- C_1 = 10 times 2mathrm~pF = 20mathrm~pF$C_1 = 10 \times 2\mathrm{~pF} = 20\mathrm{~pF}$
- C_2 = 6 times 2mathrm~pF = 12mathrm~pF$C_2 = 6 \times 2\mathrm{~pF} = 12\mathrm{~pF}$
Calculate their series equivalent C_texteq$C_{\text{eq}}$:
C_texteq = fracC_1 C_2C_1 + C_2 = frac20 times 1220 + 12 = frac24032 = 7.5mathrm~pF$$C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{20 \times 12}{20 + 12} = \frac{240}{32} = 7.5\mathrm{~pF}$$
### Step 2: Add Parallel Capacitor
Another capacitor of C_p = 7.5mathrm~pF$C_p = 7.5\mathrm{~pF}$ is connected in parallel with the combination:
C_textfinal = C_texteq + C_p = 7.5mathrm~pF + 7.5mathrm~pF = 15mathrm~pF$$C_{\text{final}} = C_{\text{eq}} + C_p = 7.5\mathrm{~pF} + 7.5\mathrm{~pF} = 15\mathrm{~pF}$$
### Pattern Recognition
Sees: Dielectric boundary parallel to the plates → Series capacitors.
Trap: Don't treat horizontally split layers as parallel; split in distance d$d$ means series, while split in area A$A$ means parallel.
Shortcut: Notice frac35.41.77$\frac{35.4}{1.77}$ is exactly 20$20$. This makes the numerical calculations incredibly clean! ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q3jee_main_2025_29_jan_eveningElectric Dipole in Uniform Electric Field
An electric dipole is placed at a distance of 2mathrm~cm$2\mathrm{~cm}$ from an infinite plane sheet having positive charge density sigma_0$\sigma_0$. Choose the correct option from the following.
The diagram illustrates an electric dipole with charges -q and +q aligned parallel to an infinite plane sheet of positive charge density.
A.textTorque on dipole is zero and net force is directed away from the sheet.$\text{Torque on dipole is zero and net force is directed away from the sheet.}$
B.textTorque on dipole is zero and net force acts towards the sheet.$\text{Torque on dipole is zero and net force acts towards the sheet.}$
C.textPotential energy of dipole is minimum and torque is zero.$\text{Potential energy of dipole is minimum and torque is zero.}$
D.textPotential energy and torque both are maximum$\text{Potential energy and torque both are maximum}$
Solution
### Related Formula
E = fracsigma_02epsilon_0$$E = \frac{\sigma_0}{2\epsilon_0}$$vectau = vecp times vecE$$\vec{\tau} = \vec{p} \times \vec{E}$$U = -vecp cdot vecE$$U = -\vec{p} \cdot \vec{E}$$
### Core Logic
An infinite plane sheet produces a uniform electric field vecE$\vec{E}$ directed normally away from the sheet.
As shown in the image layout, the dipole moment vector vecp$\vec{p}$ (pointing from -q$-q$ to +q$+q$) is oriented parallel to the electric field vectors vecE$\vec{E}$:
theta = 0^circ$$\theta = 0^{\circ}$$
1. **Torque evaluation**:
tau = pE sin(0^circ) = 0$$\tau = pE \sin(0^{\circ}) = 0$$
2. **Potential Energy evaluation**:
U = -pE cos(0^circ) = -pE quad (textMinimum)$$U = -pE \cos(0^{\circ}) = -pE \quad (\text{Minimum})$$
3. **Net Force evaluation**:
Since the electric field is uniform, the force on +q$+q$ balances the force on -q$-q$, meaning F_textnet = 0$F_{\text{net}} = 0$.
### Pattern Recognition
When a dipole aligns perfectly with a uniform electric field (vecp parallel vecE$\vec{p} \parallel \vec{E}$), it reaches stable equilibrium. Stable equilibrium fundamentally means minimum potential energy (U = -pE$U = -pE$) and zero torque.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q5jee_main_2025_29_jan_eveningGauss's Law and Flux
A point charge causes an electric flux of -2 times 10^4 mathrm~Nm^2mathrmC^-1$-2 \times 10^{4} \mathrm{~Nm}^{2}\mathrm{C}^{-1}$ to pass through a spherical Gaussian surface of 8.0mathrm~cm$8.0\mathrm{~cm}$ radius, centred on the charge. The value of the point charge is :
(mathrmGiven epsilon_0 = 8.85 times 10^-12 mathrm~C^2mathrmN^-1mathrmm^-2)$(\mathrm{Given } \epsilon_{0} = 8.85 \times 10^{-12} \mathrm{~C}^{2}\mathrm{N}^{-1}\mathrm{m}^{-2})$
A.-17.7 times 10^-8 mathrm~C$-17.7 \times 10^{-8} \mathrm{~C}$
B.-15.7 times 10^-8 mathrm~C$-15.7 \times 10^{-8} \mathrm{~C}$
C.17.7 times 10^-8 mathrm~C$17.7 \times 10^{-8} \mathrm{~C}$
D.15.7 times 10^-8 mathrm~C$15.7 \times 10^{-8} \mathrm{~C}$
Solution
### Related Formula
phi = fracq_textenclosedepsilon_0$$\phi = \frac{q_{\text{enclosed}}}{\epsilon_0}$$
where,
phi$\phi$ = net electric flux through the closed surface
q_textenclosed$q_{\text{enclosed}}$ = net charge enclosed by the surface
epsilon_0$\epsilon_0$ = permittivity of free space
### Core Logic
According to Gauss's Law, the total electric flux through a closed surface depends only on the charge enclosed inside it, completely independent of the radius of the surface.
Rearranging the formula to solve for q$q$:
q = phi cdot epsilon_0$$q = \phi \cdot \epsilon_0$$
Substitute the given values:
q = (-2 times 10^4 mathrm~Nm^2mathrmC^-1) times (8.85 times 10^-12 mathrm~C^2mathrmN^-1mathrmm^-2)$$q = (-2 \times 10^{4} \mathrm{~Nm}^2\mathrm{C}^{-1}) \times (8.85 \times 10^{-12} \mathrm{~C}^2\mathrm{N}^{-1}\mathrm{m}^{-2})$$q = -17.7 times 10^-8 mathrm~C$$q = -17.7 \times 10^{-8} \mathrm{~C}$$
### Pattern Recognition
Distractor alert: The radius (8.0mathrm~cm$8.0\mathrm{~cm}$) is extra data meant to mislead. Gauss's flux depends entirely on the magnitude of the enclosed charge, not the physical surface area configuration.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q8jee_main_2025_03_april_morningCapacitor with Multiple Dielectrics
A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant varepsilon_1$\varepsilon_1$ and varepsilon_2$\varepsilon_2$, as shown in figures. The distance between the plates is d$d$ and area of each plate is A$A$. If capacitance in first configuration and second configuration are C_1$C_1$ and C_2$C_2$ respectively, then fracC_1C_2$\frac{C_1}{C_2}$ is:
Illustrates two parallel plate configurations: stacked horizontally (series) and stacked vertically (parallel).Illustrates two parallel plate configurations: stacked horizontally (series) and stacked vertically (parallel).
### Related Formula
Capacitance with dielectric:
C = fracvarepsilon_r varepsilon_0 Ad$$C = \frac{\varepsilon_r \varepsilon_0 A}{d}$$
Series Capacitors:
C_texteq = fracC_a C_bC_a + C_b$$C_{\text{eq}} = \frac{C_a C_b}{C_a + C_b}$$
Parallel Capacitors:
C_texteq = C_a + C_b$$C_{\text{eq}} = C_a + C_b$$
### Core Logic
Let C_0 = fracvarepsilon_0 Ad$C_0 = \frac{\varepsilon_0 A}{d}$ be the capacitance without any dielectric.
- **First Configuration (Series connection)**:
The dielectrics split the gap vertically, so the effective thickness of each slab is d/2$d/2$, and the area remains A$A$.
C_a = fracvarepsilon_1 varepsilon_0 Ad/2 = 2varepsilon_1 C_0$$C_a = \frac{\varepsilon_1 \varepsilon_0 A}{d/2} = 2\varepsilon_1 C_0$$C_b = fracvarepsilon_2 varepsilon_0 Ad/2 = 2varepsilon_2 C_0$$C_b = \frac{\varepsilon_2 \varepsilon_0 A}{d/2} = 2\varepsilon_2 C_0$$
Since they are in series:
C_1 = fracC_a C_bC_a + C_b = frac(2varepsilon_1 C_0)(2varepsilon_2 C_0)2varepsilon_1 C_0 + 2varepsilon_2 C_0 = frac4varepsilon_1varepsilon_2 C_0^22C_0(varepsilon_1 + varepsilon_2) = frac2varepsilon_1varepsilon_2varepsilon_1 + varepsilon_2 C_0$$C_1 = \frac{C_a C_b}{C_a + C_b} = \frac{(2\varepsilon_1 C_0)(2\varepsilon_2 C_0)}{2\varepsilon_1 C_0 + 2\varepsilon_2 C_0} = \frac{4\varepsilon_1\varepsilon_2 C_0^2}{2C_0(\varepsilon_1 + \varepsilon_2)} = \frac{2\varepsilon_1\varepsilon_2}{\varepsilon_1 + \varepsilon_2} C_0$$
- **Second Configuration (Parallel connection)**:
The dielectrics split the area horizontally, so the effective area of each slab is A/2$A/2$, and the distance remains d$d$.
C_c = fracvarepsilon_1 varepsilon_0 (A/2)d = fracvarepsilon_1 C_02$$C_c = \frac{\varepsilon_1 \varepsilon_0 (A/2)}{d} = \frac{\varepsilon_1 C_0}{2}$$C_d = fracvarepsilon_2 varepsilon_0 (A/2)d = fracvarepsilon_2 C_02$$C_d = \frac{\varepsilon_2 \varepsilon_0 (A/2)}{d} = \frac{\varepsilon_2 C_0}{2}$$
Since they are in parallel:
C_2 = C_c + C_d = (varepsilon_1 + varepsilon_2) fracC_02$$C_2 = C_c + C_d = (\varepsilon_1 + \varepsilon_2) \frac{C_0}{2}$$Illustrates two parallel plate configurations: stacked horizontally (series) and stacked vertically (parallel).Illustrates two parallel plate configurations: stacked horizontally (series) and stacked vertically (parallel).
### Step 1: Calculating the Ratio
Now, compute fracC_1C_2$\frac{C_1}{C_2}$:
fracC_1C_2 = fracleft(frac2varepsilon_1varepsilon_2varepsilon_1 + varepsilon_2right) C_0left(fracvarepsilon_1 + varepsilon_22right) C_0 = frac4varepsilon_1varepsilon_2(varepsilon_1 + varepsilon_2)^2$$\frac{C_1}{C_2} = \frac{\left(\frac{2\varepsilon_1\varepsilon_2}{\varepsilon_1 + \varepsilon_2}\right) C_0}{\left(\frac{\varepsilon_1 + \varepsilon_2}{2}\right) C_0} = \frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1 + \varepsilon_2)^2}$$
### Pattern Recognition
For dielectric-filled capacitors: splitting the gap (d/2$d/2$) leads to a series combination, while splitting the plate area (A/2$A/2$) leads to a parallel combination.
Shortcut: C_textseries = textharmonic mean$C_{\text{series}} = \text{harmonic mean}$, C_textparallel = textarithmetic mean$C_{\text{parallel}} = \text{arithmetic mean}$. The ratio fracC_1C_2$\frac{C_1}{C_2}$ is always the ratio of the harmonic mean of the dielectric constants to their arithmetic mean!
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
Q14jee_main_2025_03_april_morningPotential of a Charged Spherical Shell
The electrostatic potential on the surface of uniformly charged spherical shell of radius R = 10mathrm~cm$R = 10\mathrm{~cm}$ is 120mathrm~V$120\mathrm{~V}$. The potential at the centre of shell, at a distance r = 5mathrm~cm$r = 5\mathrm{~cm}$ from centre, and at a distance r = 15mathrm~cm$r = 15\mathrm{~cm}$ from the centre of the shell respectively, are:
### Related Formula
For a uniformly charged spherical shell of radius R$R$ and charge Q$Q$:
- Inside and on the surface of the shell (r le R$r \le R$):
V_textin = V_textsurface = frackQR$$V_{\text{in}} = V_{\text{surface}} = \frac{kQ}{R}$$
- Outside the shell (r > R$r > R$):
V_textout = frackQr = V_textsurface left(fracRrright)$$V_{\text{out}} = \frac{kQ}{r} = V_{\text{surface}} \left(\frac{R}{r}\right)$$
### Core Logic
Let's calculate the potentials at the specified positions:
- Given surface potential at R = 10mathrm~cm$R = 10\mathrm{~cm}$ is 120mathrm~V$120\mathrm{~V}$.
1. **At the center (r = 0$r = 0$)**:
Since the center lies inside the shell (0 < 10mathrm~cm$0 < 10\mathrm{~cm}$), the potential equals the surface potential:
V_textcentre = 120mathrm~V$$V_{\text{centre}} = 120\mathrm{~V}$$
2. **At r = 5mathrm~cm$r = 5\mathrm{~cm}$**:
Since 5mathrm~cm$5\mathrm{~cm}$ is also inside the shell (5 < 10mathrm~cm$5 < 10\mathrm{~cm}$), the potential remains constant at the surface value:
V_r=5 = 120mathrm~V$$V_{r=5} = 120\mathrm{~V}$$
3. **At r = 15mathrm~cm$r = 15\mathrm{~cm}$**:
Since 15mathrm~cm$15\mathrm{~cm}$ is outside the shell (15 > 10mathrm~cm$15 > 10\mathrm{~cm}$), the potential decreases inversely with distance:
V_r=15 = V_textsurface left(fracRrright) = 120 times frac1015 = 80mathrm~V$$V_{r=15} = V_{\text{surface}} \left(\frac{R}{r}\right) = 120 \times \frac{10}{15} = 80\mathrm{~V}$$
Therefore, the potentials are 120mathrm~V$120\mathrm{~V}$, 120mathrm~V$120\mathrm{~V}$, and 80mathrm~V$80\mathrm{~V}$ respectively.
### Pattern Recognition
The electric field inside a uniformly charged conducting spherical shell is zero, meaning that no work is done moving a charge inside it. Consequently, the potential remains absolutely uniform/constant from the surface all the way to the center!
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
More Electrostatics Questions — jee_main_2026_21_jan_morning
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