Electrostatics Previous Year Questions — JEE Main Physics
79 past-year Electrostatics questions from JEE Main (Physics).
Q1 (2025)
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:
- $(A)\text{ is correct but }(R)\text{ is not correct}$
- $\text{Both }(A)\text{ and }(R)\text{ are correct but }(R)\text{ is not the correct explanation of }(A)$
- $\text{Both }(A)\text{ and }(R)\text{ are correct and }(R)\text{ is the correct explanation of }(A)$
- $(A)\text{ is not correct but }(R)\text{ is correct}$
### Related Formula
$$\vec{P}_{\text{net}} = \sum \vec{p}_i$$
where:
$\vec{P}_{\text{net}}$ = net dipole moment of the dielectric
$\vec{p}_i$ = dipole moment of the individual $i$-th molecule
### Core Logic
No external electric field is present ($E_{\text{ext}} = 0$). Due to thermal agitation, all molecular permanent dipoles are randomly oriented in space:
$$\vec{P}_{\text{net}} = 0 \quad \text{when } \vec{E}_{\text{ext}} = 0$$
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
Q9 (2025)
Two large plane parallel conducting plates are kept $10\mathrm{cm}$ apart as shown in figure. The potential difference between them is V. The potential difference between the points A and B (shown in the figure) is :
{{IMG1}}
- $\frac{1}{4} \mathrm{~V}$
- $\frac{2}{5} \mathrm{~V}$
- $\frac{3}{4} \mathrm{~V}$
- $1 \mathrm{~V}$
### Related Formula
1. Uniform Electric Field between parallel plates:
$$E = \frac{V}{d}$$
2. Potential difference between two points along the direction of the field:
$$\Delta V = E \cdot \Delta x$$
### Core Logic
The separation between the plates is $d = 10 \ \mathrm{cm}$. The electric field $\vec{E}$ between them is uniform and runs perpendicularly between the plates (along the horizontal direction, $+x$):
$$E = \frac{V}{10} \ \mathrm{V/cm}$$
From the geometry shown in the figure:
- $AC = 3 \ \mathrm{cm}$ is perpendicular to $\vec{E}$ (vertical direction).
- $CB = 4 \ \mathrm{cm}$ is parallel to $\vec{E}$ (horizontal direction).
### Step 1: Calculate potential difference between A and B
Since line $AC$ is perpendicular to $\vec{E}$:
$$V_A - V_C = 0 \implies V_A = V_C$$
Thus, the potential difference between $A$ and $B$ is completely due to the horizontal displacement $CB$:
$$V_{AB} = V_A - V_B = E \cdot (CB)$$
Substitute the values:
$$V_{AB} = \left(\frac{V}{10}\right) \times 4 = \frac{2}{5} V$$
Thus, the potential difference between $A$ and $B$ is $\frac{2}{5}V$.
### Pattern Recognition
Sees: Uniform electric field, potential difference over diagonal paths.
Trap: Projecting along the hypotenuse length ($5\text{ cm}$) directly without considering the electric field's physical direction.
Shortcut: Electric field is purely horizontal. Hence, only horizontal displacement matters. The horizontal displacement is $4\text{ cm}$ out of $10\text{ cm}$ total plate gap, so the potential difference is $\frac{4}{10}V = \frac{2}{5}V$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
Q20 (2025)
Consider a circular loop that is uniformly charged and has a radius $a\sqrt{2}$ . Find the position along the positive $z$ -axis of the cartesian coordinate system where the electric field is maximum if the ring was assumed to be placed in xy-plane at the origin:
- $\frac{a}{\sqrt{2}}$
- $\frac{a}{2}$
- $a$
- $0$
### Related Formula
1. Electric field intensity on the axis of a ring of radius $R$ at axial distance $z$:
$$E = \frac{k Q z}{\left(z^2 + R^2\right)^{3/2}}$$
2. Condition for maximum electric field on the axis of a ring:
$$z = \frac{R}{\sqrt{2}}$$
### Core Logic
To find where the axial field is maximized, take the first derivative of $E$ with respect to $z$ and set it to zero:
$$\frac{dE}{dz} = 0 \implies z^2 + R^2 - 3z^2 = 0 \implies z = \frac{R}{\sqrt{2}}$$
We are given:
- Radius of the loop $R = a\sqrt{2}$
### Step 1: Calculate the peak coordinate
Substitute the radius $R = a\sqrt{2}$ into the maximized coordinate condition:
$$z = \frac{a\sqrt{2}}{\sqrt{2}} = a$$
Thus, the electric field reaches its peak intensity at $z = a$.
### Pattern Recognition
Sees: Maximum axial field location of circular charged ring.
Trap: Simply picking radius $R$ directly as the distance value, or omitting the $1/\sqrt{2}$ scaling factor.
Shortcut: The peak of the axial field distribution of any ring always lies at $z = \frac{R}{\sqrt{2}}$. With $R = a\sqrt{2}$, $z$ resolves simply to $\frac{a\sqrt{2}}{\sqrt{2}} = a$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electric Charges and Fields
Q94 (2025)
A small bob of mass $100\mathrm{~mg}$ and charge $+10\mathrm{~\mu C}$ is connected to an insulating string of length $1\mathrm{~m}$. It is brought near to an infinitely long non-conducting sheet of charge density '$\sigma$' as shown in figure. If string subtends an angle of $45^{circ}$ with the sheet at equilibrium the charge density of sheet will be:
(Given, $\epsilon_0 = 8.85\times 10^{-12}\frac{\mathrm{F}}{\mathrm{m}}$ and acceleration due to gravity, $g = 10\mathrm{~m/s^2}$)
{{IMG1}}
- 0.885 $\mathrm{nC} / \mathrm{m}^{2}$
- 17.7 $\mathrm{nC} / \mathrm{m}^{2}$
- 885 $\mathrm{nC} / \mathrm{m}^{2}$
- 1.77 $\mathrm{nC} / \mathrm{m}^{2}$
### Related Formula
$$E = \frac{\sigma}{2\epsilon_0} \quad \text{(field of infinite non-conducting charged sheet)}$$
$$\tan\theta = \frac{F_e}{mg}$$
### Core Logic
In equilibrium, three forces act on the suspended charged bob:
1. Tension $T$ directed along the string at $\theta = 45^{\circ}$ with the vertical sheet.
2. Weight $mg$ directed vertically downwards.
3. Electrostatic repulsion force $F_e = qE$ acting horizontally away from the sheet.
From the balance of forces in vertical and horizontal directions:
$$T \cos(45^{\circ}) = mg$$
$$T \sin(45^{\circ}) = q E$$
Dividing the two equations:
$$\tan(45^{\circ}) = \frac{q E}{mg} = 1 \implies q E = mg$$
Substitute the expression for $E$:
$$q \left(\frac{\sigma}{2\epsilon_0}\right) = mg \implies \sigma = \frac{2 \epsilon_0 m g}{q}$$
Now plug in the given numerical values:
- $m = 100\mathrm{~mg} = 100 \times 10^{-6}\mathrm{~kg} = 10^{-4}\mathrm{~kg}$
- $q = +10\mathrm{~\mu C} = 10 \times 10^{-6}\mathrm{~C} = 10^{-5}\mathrm{~C}$
- $g = 10\mathrm{~m/s^2}$
- $\epsilon_0 = 8.85 \times 10^{-12}\mathrm{~F/m}$
$$\sigma = \frac{2 \times (8.85 \times 10^{-12}) \times 10^{-4} \times 10}{10^{-5}}$$
$$\sigma = 17.7 \times 10^{-10}\mathrm{~C/m}^2 = 1.77 \times 10^{-9}\mathrm{~C/m}^2 = 1.77\mathrm{~nC/m}^2$$
### Step 1: Final Conclusion
The charge density of the sheet is $1.77\mathrm{~nC/m}^2$.
### Pattern Recognition
For a charge hanging near a vertical charged sheet, the equilibrium angle is governed by $\tan\theta = \frac{F_e}{mg}$. For $\theta = 45^{\circ}$, the horizontal force equals the vertical force ($F_e = mg$). Be careful to use the field of a non-conducting sheet: $E = \frac{\sigma}{2\epsilon_0}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q5 (2025)
Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density $+\sigma$ and $-2\sigma$. The force experienced by a point charge $+q$ placed at the mid point between two plates will be:
{{IMG1}}
- $\frac{\sigma q}{4 \epsilon_0}$
- $\frac{3\sigma q}{2 \epsilon_0}$
- $\frac{3\sigma q}{4 \epsilon_0}$
- $\frac{\sigma q}{2 \epsilon_0}$
### Related Formula
$$E = \frac{\sigma_{\text{inner}}}{\epsilon_0}$$
### 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: $q_1 = \sigma$
Total charge per unit area on Plate 2: $q_2 = -2\sigma$
The outer surface charge density on the far sides of both plates must be equal:
$$\sigma_{\text{outer}} = \frac{q_1 + q_2}{2} = \frac{\sigma - 2\sigma}{2} = -\frac{\sigma}{2}$$
Now, compute the charges on the inner facing surfaces:
- Inner face of Plate 1:
$$\sigma_{\text{inner1}} = q_1 - \sigma_{\text{outer}} = \sigma - \left(-\frac{\sigma}{2}\right) = \frac{3\sigma}{2}$$
- Inner face of Plate 2:
$$\sigma_{\text{inner2}} = q_2 - \sigma_{\text{outer}} = -2\sigma - \left(-\frac{\sigma}{2}\right) = -\frac{3\sigma}{2}$$
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):
$$E = \frac{\sigma_{\text{inner1}}}{2\epsilon_0} + \frac{|\sigma_{\text{inner2}}|}{2\epsilon_0} = \frac{3\sigma/2}{2\epsilon_0} + \frac{3\sigma/2}{2\epsilon_0} = \frac{3\sigma}{2\epsilon_0}$$
Thus, the electrostatic force on $+q$ is:
$$F = q E = \frac{3\sigma q}{2\epsilon_0}$$
### Step 1: Final Conclusion
The force experienced by the point charge $+q$ is $\frac{3\sigma q}{2\epsilon_0}$.
### Pattern Recognition
For conducting plates with total charges $Q_1$ and $Q_2$, always calculate the outer charge first: $Q_{\text{outer}} = \frac{Q_1+Q_2}{2}$. The field inside the gap is exclusively due to the inner surfaces: $E = \frac{\sigma_{\text{inner}}}{\epsilon_0}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
