MHT CET 2024 10th May Evening Shift | Electrostatics Question 20 | Physics

The electric potential at a point on the axis of an electric dipole is proportional to [r = distance between centre of the electric dipole and the point]

$\frac{1}{\mathrm{r}}$

$\frac{1}{r^2}$

$r$

$\frac{1}{\mathrm{r}^3}$

MHT CET 2024 10th May Morning Shift

MCQ (Single Correct Answer)

-0

Four electric charges $+\mathrm{q},+\mathrm{q},-\mathrm{q}$ and -q are placed in order at the corners of a square of side 2 L. The electric potential at point midway between the two positive charges is

$\frac{1}{4 \pi \mathrm{E}_0} \frac{2 \mathrm{q}}{\mathrm{L}}\left(1+\frac{1}{\sqrt{5}}\right)$

$\frac{1}{4 \pi \mathrm{E}_0} \frac{2 \mathrm{q}}{L}\left(1-\frac{1}{\sqrt{5}}\right)$

$\frac{1}{4 \pi \mathrm{E}_0} \frac{2 \mathrm{q}}{\mathrm{L}}(1-\sqrt{5})$

$\frac{1}{4 \pi \mathrm{E}_0} \frac{2 \mathrm{q}}{\mathrm{L}}(1+\sqrt{5})$

MHT CET 2024 10th May Morning Shift

MCQ (Single Correct Answer)

-0

Two point charges +10 q and -4 q are located at $\mathrm{x}=0$ and $\mathrm{x}=\mathrm{L}$ respectively. What is the location of a point on the $x$-axis from the origin, which the net electric field due to these two point charges is zero?( $r=$ required distance$)$

$\mathrm{r}=\frac{\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ right to pt. B

$\mathrm{r}=\frac{\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ left to pt.A

$\mathrm{r}=\frac{\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ right to pt.B

$\mathrm{r}=\frac{\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ left to pt.A

MHT CET 2024 9th May Evening Shift

MCQ (Single Correct Answer)

-0

The van de Graaff Generator is not based on

the phenomenon of Corona Discharge in X-ray tube.

the application of electric field and magnetic field which are perpendicular to each other.

the property that charge given to a hollow conductor is transferred to its outer surface and distributed uniformly over it.

the fact that a charge is continuously supplied to an isolated metallic conductor, the potential of the conductor goes on increasing.

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