1
JEE Main 2025 (Online) 28th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A particle of mass ' $m$ ' and charge ' $q$ ' is fastened to one end ' $A$ ' of a massless string having equilibrium length $l$, whose other end is fixed at point ' $O$ '. The whole system is placed on a frictionless horizontal plane and is initially at rest. If uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the $x$-axis is

JEE Main 2025 (Online) 28th January Morning Shift Physics - Electrostatics Question 1 English

A
$\sqrt{\frac{\mathrm{qE} l}{2 \mathrm{~m}}}$
B
$\sqrt{\frac{\mathrm{qE} l}{4 \mathrm{~m}}}$
C
$\sqrt{\frac{\mathrm{qE} l}{\mathrm{~m}}}$
D
$\sqrt{\frac{2 \mathrm{qE} l}{\mathrm{~m}}}$
2
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A small uncharged conducting sphere is placed in contact with an identical sphere but having $4 \times 10^{-8} \mathrm{C}$ charge and then removed to a distance such that the force of repulsion between them is $9 \times 10^{-3} \mathrm{~N}$. The distance between them is (Take $\frac{1}{4 \pi \epsilon_{\mathrm{o}}}$ as $9 \times 10^9$ in SI units)

A
1 cm
B
2 cm
C
4 cm
D
3 cm
3
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

JEE Main 2025 (Online) 24th January Evening Shift Physics - Electrostatics Question 3 English

In the first configuration (1) as shown in the figure, four identical charges $\left(q_0\right)$ are kept at the corners A, B, C and D of square of side length ' $a$ '. In the second configuration (2), the same charges are shifted to mid points $G, E, H$ and $F$, of the square. If $K=\frac{1}{4 \pi \epsilon_0}$, the difference between the potential energies of configuration (2) and (1) is given by :

A
$\frac{K q_0^2}{a}(4-2 \sqrt{2})$
B
$\frac{K q_0^2}{a}(4 \sqrt{2}-2)$
C
$\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3-\sqrt{2})$
D
$\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3 \sqrt{2}-2)$
4
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Consider a parallel plate capacitor of area A (of each plate) and separation ' $d$ ' between the plates. If $E$ is the electric field and $\varepsilon_0$ is the permittivity of free space between the plates, then potential energy stored in the capacitor is

A
$\varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$
B
$\frac{3}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$
C
$\frac{1}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$
D
$\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}$
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