1
MHT CET 2024 15th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

In an electric field due to charge $Q$, a charge $q$ moves from point A to B as shown in the figure. The work done is ( $\varepsilon_0=$ permittivity of free space)

MHT CET 2024 15th May Evening Shift Physics - Electrostatics Question 18 English

A
$\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}^2}$
B
$\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}^2} \frac{\pi}{6}$
C
$\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{\mathrm{r}}$
D
zero
2
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If a unit positive charge is shifted from a region of low potential to a region of high potential, then the electric potential energy of the system

A
increases.
B
decreases.
C
does not change.
D
is zero.
3
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Two point charges $+8 q$ and $-2 q$ are located at $\mathrm{X}=0$ (origin) and $\mathrm{X}=\mathrm{L}$ respectively. The net electric field due to these two charges is zero at point $P$ on $X$-axis. The location of point $P$ from the origin is

A
$\frac{L}{4}$
B
2L
C
4L
D
8L
4
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Consider a long uniformly charged cylinder having constant volume charge density ' $\lambda$ ' and radius ' $R$ '. A Gaussian surface is in the form of a cylinder of radius ' $r$ ' such that vertical axis of both the cylinders coincide. For a point inside the cylinder $(r< R)$, electric field is directly proportional to

A
$\mathrm{r}^{-1}$
B
$\mathrm{r}$
C
$\mathrm{r}^2$
D
$\mathrm{r}^{-2}$
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