Electrostatics JEE Main Physics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: (c)

Explanation: The outer body of air craft made with metal to provide electrostatic shield, such that electric field inside craft remain zero for any charge on outer surface.

Correct Answer: (d)

Explanation: For $r \leq R, V=\;\frac{K Q}{R}$ , i.e., Constant everywhere inside. For $r > R, V=\;\frac{K Q}{r}$ , i.e., Decreases with $r$ .

Correct Answer: 12

Explanation: $\phi = \vec{E} \cdot \vec{A}$ $=\left(\frac{\widehat{2\hat{i}+6\hat{j}+8\hat{k}}}{\sqrt{6}}\right) \cdot 4\left(\frac{\widehat{2\hat{i}+\hat{j}+\hat{k}}}{\sqrt{6}}\right)$ $=\frac{4}{6} \times (4+6+8)=12\ \mathrm{Vm}$

Correct Answer: (d)

Explanation: Suppose the two charges $+q$ and $+q$ are separated by a distance $2a$. So the charge $Q$ is at the midpoint, at a distance $a$ from each $+q$. For the whole system to be in equilibrium, the net force on each charge must be zero. The force on the middle charge $Q$ due to the two equal charges $+q$ is already zero, because both forces are equal and opposite. Now consider one end charge $+q$. It is repelled by the other end charge $+q$. Distance between the two $+q$ charges is $2a$, so this repulsive force is $ F_1 = \frac{kq^2}{(2a)^2} $ $ F_1 = \frac{kq^2}{4a^2} $ To balance this force, the middle charge $Q$ must attract the end charge $+q$. So $Q$ must be negative. Force between $Q$ and $+q$ is $ F_2 = \frac{kq|Q|}{a^2} $ For equilibrium, $ F_1 = F_2 $ $ \frac{kq^2}{4a^2} = \frac{kq|Q|}{a^2} $ Cancel common terms: $ |Q| = \frac{q}{4} $ Since $Q$ must be negative, $ Q = -\frac{q}{4} $ So the correct option is $ \boxed{\text{Option D } \frac{-q}{4}} $

Correct Answer: (c)

Explanation: When eight small liquid drops combine to form one large drop, the potential of the large drop is given as 20 V. We need to find the potential of each individual small drop. Solution **Volume Relationship:** The total volume of the eight small drops is equal to the volume of the one large drop. This relationship is expressed as: $ \left(\frac{4}{3} \pi r^3\right) \times 8 = \frac{4}{3} \pi R^3 $ Simplifying gives: $ 2r = R \quad \ldots \text{(i)} $ This equation indicates the radius of the large drop $ R $ is twice the radius of one small drop $ r $. **Charge Conservation:** The total charge of the eight small drops is equal to the charge of the large drop: $ 8q = Q \quad \ldots \text{(ii)} $ **Potential Derivation:** The potential for one small drop ($ V' $) and the large drop ($ V $) are given by: $ V' = \frac{q}{4 \pi \varepsilon_0 r} $ $ V = \frac{Q}{4 \pi \varepsilon_0 R} $ **Finding the Ratio:** Using the above expressions, we derive: $ \frac{V'}{V} = \frac{q}{Q} \times \frac{R}{r} $ Substituting from equations (i) and (ii), we get: $ \frac{V'}{20} = \frac{q}{8q} \times \frac{2r}{r} $ This simplifies to: $ \frac{V'}{20} = \frac{1}{4} $ **Calculating $ V' $:** Solving for $ V' $, we find: $ V' = 5 \, \text{V} $ Thus, the potential of each single small liquid drop is 5 V.