1
MHT CET 2025 22nd April Evening Shift
MCQ (Single Correct Answer)
+1
-0

A charge $\mathrm{Q} \mu \mathrm{C}$ is placed at the centre of a cube. The flux through two opposite faces of the cube is ( $\varepsilon_0=$ permittivity of free space)

A
$\frac{\mathrm{Q}}{6 \varepsilon_0}$
B
$\frac{\mathrm{Q}}{3 \varepsilon_0}$
C
$\frac{\mathrm{Q}}{\varepsilon_0}$
D
$\frac{\mathrm{Q}}{2 \varepsilon_0}$
2
MHT CET 2025 22nd April Morning Shift
MCQ (Single Correct Answer)
+1
-0

Four charges $2 \mu \mathrm{C},-3 \mu \mathrm{C}, 4 \mu \mathrm{C},-4 \mu \mathrm{C}$ and $-1 \mu \mathrm{C}$ are enclosed by the Gaussian surface of radius 2 m . Net outward flux through the Gaussian surface is (in $\mu \mathrm{V}-\mathrm{m}$ ) [ $\varepsilon_0=$ permittivity of free space]

A
$\frac{2}{\varepsilon_0}$
B
zero
C
$\frac{3}{\varepsilon_0}$
D
$\frac{5}{\varepsilon_0}$
3
MHT CET 2025 21st April Evening Shift
MCQ (Single Correct Answer)
+1
-0

The electric field intensity on the surface of a solid charged sphere of radius $\mathbf{r}$ and volume charge density $\sigma$ is ( $\varepsilon_0=$ permittivity of free space)

A
zero
B
$\frac{5 \sigma r}{6 \varepsilon_0}$
C
$\frac{1}{4 \pi \varepsilon_0} \frac{\sigma}{\mathrm{r}}$
D
$\frac{\sigma \mathrm{r}}{3 \varepsilon_0}$
4
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+1
-0

The electric charges ' $+2 q$ ', ' $+2 q$ ', ' $-2 q$ ' and ' $-2 q$ ' are placed at the corners of square of side ' 2 L ' as shown in figure. The electric potential at point 'A', midway between the two charges ' $+2 q$ ' and ' $+2 q$ ' is

( $\varepsilon_0=$ permittivity of free space)

MHT CET 2025 21st April Morning Shift Physics - Electrostatics Question 17 English
A
$\quad \frac{1}{4 \pi \varepsilon_0}\left(\frac{2 q}{L}\right)\left[1+\frac{1}{\sqrt{5}}\right]$
B
$\frac{\mathrm{q}}{\pi \varepsilon_0 \mathrm{~L}}\left[1-\frac{1}{\sqrt{5}}\right]$
C
$\frac{\mathrm{q}}{\pi \varepsilon_0 \mathrm{~L}}\left[1+\frac{1}{\sqrt{5}}\right]$
D
$\frac{1}{4 \pi \varepsilon_0}\left(\frac{2 q}{L}\right)\left[1-\frac{1}{\sqrt{5}}\right]$
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