NEET 2023 Manipur | Electrostatics Question 8 | Physics

If a conducting sphere of radius $$\mathrm{R}$$ is charged. Then the electric field at a distance $$\mathrm{r}(\mathrm{r} > \mathrm{R})$$ from the centre of the sphere would be, $$(\mathrm{V}=$$ potential on the surface of the sphere)

$$\frac{r V}{R^2}$$

$$\frac{R^2 V}{r^3}$$

$$\frac{R V}{r^2}$$

$$\frac{\mathrm{V}}{\mathrm{r}}$$

NEET 2023

MCQ (Single Correct Answer)

-1

An electric dipole is placed at an angle of $$30^{\circ}$$ with an electric field of intensity $$2 \times 10^{5} \mathrm{NC}^{-1}$$. It experiences a torque equal to $$4 ~\mathrm{N~m}$$. Calculate the magnitude of charge on the dipole, if the dipole length is $$2 \mathrm{~cm}$$.

6 mC

4 mC

2 mC

8 mC

NEET 2023

MCQ (Single Correct Answer)

-1

If $$\oint_\limits{s} \vec{E} \cdot \overrightarrow{d S}=0$$ over a surface, then:

the magnitude of electric field on the surface is constant.

all the charges must necessarily be inside the surface.

the electric field inside the surface is necessarily uniform.

the number of flux lines entering the surface must be equal to the number of flux lines leaving it.

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