1
JEE Main 2025 (Online) 2nd April Morning Shift
MCQ (Single Correct Answer)
+4
-1

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:

JEE Main 2025 (Online) 2nd April Morning Shift Physics - Electrostatics Question 8 English

A
$\frac{3 \sigma q}{4 \epsilon_0}$
B
$\frac{3 \sigma \mathrm{q}}{2 \epsilon_0}$
C
$\frac{\sigma \mathrm{q}}{4 \epsilon_0}$
D
$\frac{\sigma q}{2 \epsilon_0}$
2
JEE Main 2025 (Online) 2nd April Morning Shift
MCQ (Single Correct Answer)
+4
-1

A point charge $+q$ is placed at the origin. A second point charge $+9 q$ is placed at ($\mathrm{d}, 0,0$) in Cartesian coordinate system. The point in between them where the electric field vanishes is:

A
$(3 \mathrm{d} / 4,0,0)$
B
$(\mathrm{d} / 4,0,0)$
C
$(4 \mathrm{d} / 3,0,0)$
D
$(\mathrm{d} / 3,0,0)$
3
JEE Main 2025 (Online) 2nd April Morning Shift
MCQ (Single Correct Answer)
+4
-1

A small bob of mass 100 mg and charge $+10 \mu \mathrm{C}$ is connected to an insulating string of length 1 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, $\mathrm{g}=10 \frac{\mathrm{~m}}{\mathrm{~s}^2}$ )

JEE Main 2025 (Online) 2nd April Morning Shift Physics - Electrostatics Question 7 English

A
$1.77 \mathrm{~nC} / \mathrm{m}^2$
B
$0.885 \mathrm{~nC} / \mathrm{m}^2$
C
$885 \mathrm{~nC} / \mathrm{m}^2$
D
$17.7 \mathrm{~nC} / \mathrm{m}^2$
4
JEE Main 2025 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A point charge causes an electric flux of $-2 \times 10^4 \mathrm{Nm}^2 \mathrm{C}^{-1}$ to pass through a spherical Gaussian surface of 8.0 cm radius, centred on the charge. The value of the point charge is :

(Given $\epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$ )

A
$17.7 \times 10^{-8} \mathrm{C}$
B
$-17.7 \times 10^{-8} \mathrm{C}$
C
$15.7 \times 10^{-8} \mathrm{C}$
D
$-15.7 \times 10^{-8} \mathrm{C}$
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