Electrostatics Previous Year Questions — JEE Main Physics
79 past-year Electrostatics questions from JEE Main (Physics) — page 10.
Q1625 (2024)
Two charges of $5\mathrm{Q}$ and $-2\mathrm{Q}$ are situated at the points $(3a, 0)$ and $(-5a, 0)$ respectively. The electric flux through a sphere of radius '4a' having center at origin is:
- $\frac{2Q}{\varepsilon_0}$
- $\frac{5Q}{\varepsilon_0}$
- $\frac{7 \mathrm{Q}}{\varepsilon_{0}}$
- $\frac{3Q}{\varepsilon_0}$
### Related Formula
Gauss's Law states:
$$\Phi = \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$
### Core Logic
We have a sphere of radius $R = 4a$ centered at $(0,0)$.
* Charge $5Q$ is located at $(3a, 0)$. Since the distance from origin is $3a \lt 4a$, this charge lies **inside** the sphere.
* Charge $-2Q$ is located at $(-5a, 0)$. Since the distance from origin is $5a \gt 4a$, this charge lies **outside** the sphere.
{{SOL_IMG1}}
### Step 1: Calculate Enclosed Charge
The net enclosed charge $q_{\text{enclosed}}$ within the spherical boundary is:
$$q_{\text{enclosed}} = 5Q$$
### Step 2: Apply Gauss Law
Using Gauss's Law, the total electric flux is:
$$\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{5Q}{\varepsilon_0}$$
Therefore, the electric flux is $\frac{5Q}{\varepsilon_0}$.
### Pattern Recognition
Flux depends purely on charges situated *inside* the closed surface. Charges outside the Gaussian surface contribute absolutely zero net flux because every field line entering must also exit.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electric Charges and Fields
Q59 (2024)
An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet S having surface charge density $+\sigma$. The electron at t = 0 is at a distance of 1 m from S and has a speed of 1 m/s. The maximum value of $\sigma$ if the electron strikes S at t = 1 s is $\alpha \left[\frac{\mathrm{m} \epsilon_{0}}{\mathrm{e}}\right] \frac{\mathrm{C}}{\mathrm{m}^{2}}$ the value of $\alpha$ is
### Related Formula
The uniform electric field ($E$) produced by an infinite plane sheet with positive charge density $+\sigma$ is:
$$E = \frac{\sigma}{2\varepsilon_0}$$
The constant electrostatic force acting on an electron (charge $-e$, mass $m$) directed towards the sheet is:
$$F = e E = \frac{\sigma e}{2\varepsilon_0} \implies \text{Acceleration } a = -\frac{\sigma e}{2\varepsilon_0 m}$$$
### Core Logic
For the boundary limits matching a successful collision with the plate under maximum field conditions, the electron must be moving directly away from sheet $S$ initially ($u = +1 \mathrm{~m/s}$). The accelerating field acts as a decelerating force, turning it around to hit the sheet exactly at $t = 1 \mathrm{~s}$.
Therefore, relative to the initial position vector pointing outwards:
* Initial speed $u = 1 \mathrm{~m/s}$
* Net displacement at strike $S = -1 \mathrm{~m}$ (since it returns to the plate at origin)
* Total time elapsed $t = 1 \mathrm{~s}$
### Step 1: Apply Second Equation of Motion
Using $S = u t + \frac{1}{2} a t^2$:
$$-1 = 1 \times 1 + \frac{1}{2} a (1)^2$$
$$-1 = 1 + \frac{1}{2} a \implies \frac{1}{2} a = -2 \implies a = -4 \mathrm{~m/s^2}$$
### Step 2: Solve for Surface Charge Density
Equating this deceleration value to the electrostatic tracking acceleration:
$$-4 = -\frac{\sigma e}{2\varepsilon_0 m}$$
$$\sigma = 8 \frac{\varepsilon_0 m}{e} = 8 \left[ \frac{m \varepsilon_0}{e} \right]$$
### Step 3: Extract alpha
Comparing this with the algebraic pattern $\alpha \left[\frac{m \varepsilon_0}{e}\right]$:
$\alpha = 8$
Therefore, the value of $\alpha$ is $8$.
### Pattern Recognition
Be careful with coordinate signs. If the electron were moving towards the plate initially ($u = -1$), any field value would pull it in even faster, meaning the time would be less than $1 \mathrm{~s}$. The 'maximum field' boundary constraint implies the electron is thrown away and turned around at its apex peak, mapping perfectly to a return displacement.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electric Charges and Fields
Class 11 Physics: Motion in a Straight Line
Q1737 (2024)
Two identical charged spheres are suspended by string of equal lengths. The string makes an angle of $37^{\circ}$ with each other. When suspended in a liquid of density $0.7 \mathrm{~g/cm}^3$, the angle remains same. If density of material of the sphere is $1.4 \mathrm{~g/cm}^3$, the dielectric constant of the liquid is ($\tan 37^{\circ} = \frac{3}{4}$).
### Related Formula
$$\tan \theta = \frac{F_e}{mg}$$
$$F_e' = \frac{F_e}{k}$$
$$W_{\text{apparent}} = mg - F_b = V (\rho_B - \rho_L) g$$
### Core Logic
{{SOL_IMG1}}
For a charged sphere suspended in air, the equilibrium condition gives:
$$T \cos\theta = mg$$
$$T \sin\theta = F_e$$
$$\tan\theta = \frac{F_e}{mg} = \frac{F_e}{\rho_B V g} \quad \dots (i)$$
When suspended in a liquid, both the electrostatic force and the effective weight change.
The new electrostatic force is $F_e' = \frac{F_e}{k}$, where $k$ is the dielectric constant.
The apparent weight is $W' = V \rho_B g - V \rho_L g = V(\rho_B - \rho_L)g$.
Since the angle remains the same, $\tan\theta$ is unchanged:
$$\tan\theta = \frac{F_e'}{W'} = \frac{F_e / k}{V(\rho_B - \rho_L)g} \quad \dots (ii)$$
### Step 1: Equate and Solve for k
Equating (i) and (ii):
$$\frac{F_e}{\rho_B V g} = \frac{F_e}{k V (\rho_B - \rho_L) g}$$
$$\rho_B = k (\rho_B - \rho_L)$$
Substitute the given densities:
$\rho_B = 1.4 \mathrm{~g/cm}^3$
$\rho_L = 0.7 \mathrm{~g/cm}^3$
$$1.4 = k (1.4 - 0.7)$$
$$1.4 = 0.7 k \implies k = 2$$
### Pattern Recognition
For this classic setup where the angle remains unaltered in a dielectric liquid, the dielectric constant formula is strictly $k = \frac{\rho_{\text{body}}}{\rho_{\text{body}} - \rho_{\text{liquid}}}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electrostatics
Class 11 Physics: Mechanical Properties of Fluids
Q43 (2024)
A particle of charge $-q$ and mass $m$ moves in a circle of radius $r$ around an infinitely long line charge of linear density $+ \lambda$. Then time period will be given as:
(Consider k as Coulomb's constant)
- $\mathrm{T}^{2} = \frac{4\pi^{2}\mathrm{m}}{2\mathrm{k}\lambda\mathrm{q}}\mathrm{r}^{3}$
- $\mathrm{T} = 2\pi \mathrm{r}\sqrt{\frac{\mathrm{m}}{2\mathrm{k}\lambda\mathrm{q}}}$
- $\mathrm{T} = \frac{1}{2\pi\mathrm{r}}\sqrt{\frac{\mathrm{m}}{2\mathrm{k}\lambda\mathrm{q}}}$
- $\mathrm{T} = \frac{1}{2\pi}\sqrt{\frac{2\mathrm{k}\lambda\mathrm{q}}{\mathrm{m}}}$
### Related Formula
$$E = \frac{2k\lambda}{r}$$
$$F_c = m\omega^2 r = \frac{mv^2}{r}$$
### Core Logic
For circular motion, the required centripetal force is provided by the electrostatic force of attraction between the negatively charged particle and the positively charged infinite line charge.
$$F_e = qE = q \left(\frac{2k\lambda}{r}\right)$$
Equating this to the centripetal force $m\omega^2 r$:
### Step 1: Solve for Angular Velocity
$$\frac{2k\lambda q}{r} = m\omega^2 r$$
$$\omega^2 = \frac{2k\lambda q}{mr^2}$$
### Step 2: Solve for Time Period
Since $T = \frac{2\pi}{\omega}$:
$$\left(\frac{2\pi}{T}\right)^2 = \frac{2k\lambda q}{mr^2}$$
$$\frac{2\pi}{T} = \sqrt{\frac{2k\lambda q}{mr^2}}$$
$$T = 2\pi r \sqrt{\frac{m}{2k\lambda q}}$$
### Pattern Recognition
When a particle orbits a line charge, the electrostatic force scales as $1/r$. The centripetal force $m v^2/r$ means $v$ is independent of $r$. Hence, the time period $T = 2\pi r / v$ is directly proportional to $r$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Class 11 Physics: Laws of Motion
Q46 (2024)
The electrostatic potential due to an electric dipole at a distance $r$ varies as:
- $r$
- $\frac{1}{r^2}$
- $\frac{1}{r^3}$
- $\frac{1}{r}$
### Related Formula
$$V = \frac{1}{4\pi\varepsilon_0} \frac{p \cos \theta}{r^2}$$
### Core Logic
For a short electric dipole, the potential $V$ at a general point $(r, \theta)$ is inversely proportional to the square of the distance from the center of the dipole.
### Step 1: Final Conclusion
From the formula $V = \frac{k p \cos \theta}{r^2}$, it is evident that $V \propto \frac{1}{r^2}$.
### Pattern Recognition
Point charge potential $\propto 1/r$. Dipole potential falls off faster $\propto 1/r^2$. Quadrupole potential falls off $\propto 1/r^3$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
