Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Net dipole moment of a polar linear isotropic dielectric substance is not zero even in the absence of an external electric field.
Reason (R): In absence of an external electric field, the different permanent dipoles of a polar dielectric substance are oriented in random directions.
In the light of the above statements, choose the most appropriate answer from the options given below:
A.(A)text is correct but (R)text is not correct$(A)\text{ is correct but }(R)\text{ is not correct}$
B.textBoth (A)text and (R)text are correct but (R)text is not the correct explanation of (A)$\text{Both }(A)\text{ and }(R)\text{ are correct but }(R)\text{ is not the correct explanation of }(A)$
C.textBoth (A)text and (R)text are correct and (R)text is the correct explanation of (A)$\text{Both }(A)\text{ and }(R)\text{ are correct and }(R)\text{ is the correct explanation of }(A)$
D.(A)text is not correct but (R)text is correct$(A)\text{ is not correct but }(R)\text{ is correct}$
Solution & Explanation
### Related Formula
vecP_textnet = sum vecp_i$$\vec{P}_{\text{net}} = \sum \vec{p}_i$$
where:
vecP_textnet$\vec{P}_{\text{net}}$ = net dipole moment of the dielectric
vecp_i$\vec{p}_i$ = dipole moment of the individual i$i$-th molecule
### Core Logic
No external electric field is present (E_textext = 0$E_{\text{ext}} = 0$). Due to thermal agitation, all molecular permanent dipoles are randomly oriented in space:
vecP_textnet = 0 quad textwhen vecE_textext = 0$$\vec{P}_{\text{net}} = 0 \quad \text{when } \vec{E}_{\text{ext}} = 0$$
Thus:
1. Assertion (A) is false because it claims the net dipole moment is non-zero even without an external field.
2. Reason (R) is true because it correctly describes that different permanent dipoles are randomly oriented.
### Step 1: Final Conclusion
Therefore, (A) is not correct but (R) is correct.
### Pattern Recognition
Sees: "polar dielectric" + "no external field" → net bulk dipole moment is always zero.
Trap: Confusing the molecular level with the macroscopic level. Each molecule in a polar dielectric has a permanent dipole moment, but the macro substance has zero net moment due to random thermal orientations.
Shortcut: No external field means vectors cancel globally, which implies zero net moment. Thus (A) is false immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Keywords:#Net dipole moment of a polar linear isotropic dielectric#JEE Main 2025 Evening Q1#Electrostatics JEE Main 2025#Dielectrics and Polarization JEE Main 2025
More Electrostatics Previous-Year Questions
Qjee_main_2026_21_jan_morningCapacitance with Dielectric
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
Q45jee_main_2026_21_jan_morningElectric Potential Energy
A point charge of 10^-8$10^{-8}$ C is placed at origin. The work done in moving a point charge 2 mu$\mu$C from point A(4, 4, 2) m to B(2, 2, 1) m is ____ J. left(frac14piepsilon_0=9times10^9text in SI unitsright)$\left(\frac{1}{4\pi\epsilon_{0}}=9\times10^{9}\text{ in SI units}\right)$
A.45 times 10^-6$45 \times 10^{-6}$
B.0$0$
C.30 times 10^-6$30 \times 10^{-6}$
D.15 times 10^-6$15 \times 10^{-6}$
Solution
### Related Formula
W_textext = Delta U = U_f - U_i$$W_{\text{ext}} = \Delta U = U_f - U_i$$U = frac14piepsilon_0 fracq_1 q_2r$$U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$$
### Core Logic
Work done by external agent: W_textext = Delta U$W_{\text{ext}} = \Delta U$, where Delta U$\Delta U$ is the change in potential energy.
W_textext = frac14pi epsilon_0 fracq_1 q_2r_f - frac14pi epsilon_0 fracq_1 q_2r_i$$W_{\text{ext}} = \frac{1}{4\pi \epsilon_{0}} \frac{q_{1} q_{2}}{r_{f}} - \frac{1}{4\pi \epsilon_{0}} \frac{q_{1} q_{2}}{r_{i}}$$
Calculate the distances of points A and B from the origin:
r_i = |A| = sqrt4^2 + 4^2 + 2^2 = sqrt16+16+4 = sqrt36 = 6text m$r_i = |A| = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16+16+4} = \sqrt{36} = 6\text{ m}$r_f = |B| = sqrt2^2 + 2^2 + 1^2 = sqrt4+4+1 = sqrt9 = 3text m$r_f = |B| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3\text{ m}$
### Step 1: Calculate Work Done
W_textext = (9 times 10^9) times (10^-8 times 2 times 10^-6) left[ frac13 - frac16 right]$$W_{\text{ext}} = (9 \times 10^9) \times (10^{-8} \times 2 \times 10^{-6}) \left[ \frac{1}{3} - \frac{1}{6} \right]$$W_textext = 18 times 10^-5 times left(frac2-16right)$$W_{\text{ext}} = 18 \times 10^{-5} \times \left(\frac{2-1}{6}\right)$$W_textext = 18 times 10^-5 times frac16 = 3 times 10^-5text J = 30 times 10^-6text J$$W_{\text{ext}} = 18 \times 10^{-5} \times \frac{1}{6} = 3 \times 10^{-5}\text{ J} = 30 \times 10^{-6}\text{ J}$$
### Pattern Recognition
Electric field is conservative. Work done simply equals change in kqq/r$kqq/r$ from initial to final radial coordinate. No path dependence.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q5jee_main_2025_02_april_morningElectric Field and Gauss's Law
Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density +sigma$+\sigma$ and -2sigma$-2\sigma$. The force experienced by a point charge +q$+q$ placed at the mid point between two plates will be:
Two parallel plates with charges +sigma and -2sigma and a point charge +q at the midpoint.
### Related Formula
E = fracsigma_textinnerepsilon_0$$E = \frac{\sigma_{\text{inner}}}{\epsilon_0}$$
### Core Logic
For parallel conducting plates of large area, charges redistribute on the outer and inner faces to maintain electrostatic equilibrium.
Total charge per unit area on Plate 1: q_1 = sigma$q_1 = \sigma$
Total charge per unit area on Plate 2: q_2 = -2sigma$q_2 = -2\sigma$
The outer surface charge density on the far sides of both plates must be equal:
sigma_textouter = fracq_1 + q_22 = fracsigma - 2sigma2 = -fracsigma2$$\sigma_{\text{outer}} = \frac{q_1 + q_2}{2} = \frac{\sigma - 2\sigma}{2} = -\frac{\sigma}{2}$$
Now, compute the charges on the inner facing surfaces:
- Inner face of Plate 1:
sigma_textinner1 = q_1 - sigma_textouter = sigma - left(-fracsigma2right) = frac3sigma2$$\sigma_{\text{inner1}} = q_1 - \sigma_{\text{outer}} = \sigma - \left(-\frac{\sigma}{2}\right) = \frac{3\sigma}{2}$$
- Inner face of Plate 2:
sigma_textinner2 = q_2 - sigma_textouter = -2sigma - left(-fracsigma2right) = -frac3sigma2$$\sigma_{\text{inner2}} = q_2 - \sigma_{\text{outer}} = -2\sigma - \left(-\frac{\sigma}{2}\right) = -\frac{3\sigma}{2}$$
In the region between the plates, both inner surfaces create an electric field in the same direction (away from the positive plate 1 and towards negative plate 2):
E = fracsigma_textinner12epsilon_0 + frac|sigma_textinner2|2epsilon_0 = frac3sigma/22epsilon_0 + frac3sigma/22epsilon_0 = frac3sigma2epsilon_0$$E = \frac{\sigma_{\text{inner1}}}{2\epsilon_0} + \frac{|\sigma_{\text{inner2}}|}{2\epsilon_0} = \frac{3\sigma/2}{2\epsilon_0} + \frac{3\sigma/2}{2\epsilon_0} = \frac{3\sigma}{2\epsilon_0}$$
Thus, the electrostatic force on +q$+q$ is:
F = q E = frac3sigma q2epsilon_0$$F = q E = \frac{3\sigma q}{2\epsilon_0}$$
### Step 1: Final Conclusion
The force experienced by the point charge +q$+q$ is frac3sigma q2epsilon_0$\frac{3\sigma q}{2\epsilon_0}$.
### Pattern Recognition
For conducting plates with total charges Q_1$Q_1$ and Q_2$Q_2$, always calculate the outer charge first: Q_textouter = fracQ_1+Q_22$Q_{\text{outer}} = \frac{Q_1+Q_2}{2}$. The field inside the gap is exclusively due to the inner surfaces: E = fracsigma_textinnerepsilon_0$E = \frac{\sigma_{\text{inner}}}{\epsilon_0}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q7jee_main_2025_02_april_morningElectric Field and Gauss's Law
A point charge +q$+q$ is placed at the origin. A second point charge +9q$+9q$ is placed at (d, 0, 0)$(d, 0, 0)$ in Cartesian coordinate system. The point in between them where the electric field vanishes is:
A.(4d / 3, 0, 0)$(4d / 3, 0, 0)$
B.(d / 4, 0, 0)$(d / 4, 0, 0)$
C.(3d / 4, 0, 0)$(3d / 4, 0, 0)$
D.(d / 3, 0, 0)$(d / 3, 0, 0)$
Solution
### Related Formula
E = frack Qr^2$$E = \frac{k Q}{r^2}$$x = fracd1 + sqrtfracq_2q_1$$x = \frac{d}{1 + \sqrt{\frac{q_2}{q_1}}}$$
### Core Logic
Let the null point where the electric field is zero be at (x, 0, 0)$(x, 0, 0)$ where 0 < x < d$0 < x < d$.
At this point, the fields due to both charges are equal in magnitude and opposite in direction:
frack qx^2 = frack (9q)(d - x)^2$$\frac{k q}{x^2} = \frac{k (9q)}{(d - x)^2}$$
Taking the square root on both sides:
frac1x = frac3d - x implies d - x = 3x$$\frac{1}{x} = \frac{3}{d - x} \implies d - x = 3x$$4x = d implies x = fracd4$$4x = d \implies x = \frac{d}{4}$$
Thus, the coordinates of the null point are left(fracd4, 0, 0right)$\left(\frac{d}{4}, 0, 0\right)$.
### Step 1: Final Conclusion
The point where the electric field vanishes is left(fracd4, 0, 0right)$\left(\frac{d}{4}, 0, 0\right)$.
### Pattern Recognition
For two like charges, the zero-field null point always lies along the line joining them and is closer to the smaller charge. Use the standard shortcut: x = fracd1 + sqrtq_2/q_1$x = \frac{d}{1 + \sqrt{q_2/q_1}}$ measured from charge q_1$q_1$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
More Electrostatics Questions — jee_main_2025_02_april_evening
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