MHT CET 2024 10th May Evening Shift | Electrostatics Question 19 | Physics

Two point charges $(A$ and $B)+4 q$ and $-4 q$ are placed along a line separated by a distance I '. Force acting between them is F. If $25 \%$ of charge from point $A$ is transferred to that at point B , the force between the charges now becomes

$\frac{3}{4} \mathrm{~F}$

$\frac{4}{3} \mathrm{~F}$

$\frac{9}{16} \mathrm{~F}$

$\frac{16}{9} \mathrm{~F}$

MHT CET 2024 10th May Evening Shift

MCQ (Single Correct Answer)

-0

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

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