MHT CET 2023 14th May Evening Shift | Electrostatics Question 2 | Physics

If $$\mathrm{E}_{\mathrm{a}}$$ and $$\mathrm{E}_{\mathrm{q}}$$ represent the electric field intensity due to a short dipole at a point on its axial line and on the equatorial line at the same distance '$$r$$' from the centre of the dipole, then

$$\mathrm{E}_{\mathrm{a}}=\mathrm{E}_{\mathrm{q}}$$

$$\mathrm{E}_{\mathrm{a}}=\frac{1}{2} \mathrm{E}_{\mathrm{q}}$$

$$\mathrm{E}_{\mathrm{a}}=\frac{1}{\sqrt{2}} \mathrm{E}_{\mathrm{q}}$$

$$\mathrm{E}_{\mathrm{a}}=2 \mathrm{E}_{\mathrm{q}}$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

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

$$\frac{\rho r}{3 \varepsilon_0}$$

$$\frac{\rho}{4 \pi \varepsilon_0 \mathrm{r}}$$

zero

$$\frac{5 \rho \mathrm{r}}{6 \varepsilon_0}$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

A uniformly charged semicircular arc of radius '$$r$$' has linear charge density '$$\lambda$$'. The electric field at its centre is ( $$\varepsilon_0=$$ permittivity of free space)

$$\frac{\lambda}{4 \varepsilon_0}$$

$$\frac{2 \varepsilon_0}{\lambda}$$

$$\frac{\lambda}{4 \varepsilon_0 \mathrm{r}}$$

$$\frac{2 \pi \varepsilon_0}{\lambda}$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

A conducting sphere of radius $$0.1 \mathrm{~m}$$ has uniform charge density $$1.8 \mu \mathrm{C} / \mathrm{m}^2$$ on its surface. The electric field in free space at radial distance $$0.2 \mathrm{~m}$$ from a point on the surface is ( $$\varepsilon_0=$$ permittivity of free space)

$$\frac{6 \times 10^{-6}}{\varepsilon_0} \mathrm{Vm}^{-1}$$

$$\frac{6 \times 10^{-8}}{\varepsilon_0} \mathrm{Vm}^{-1}$$

$$\frac{2 \times 10^{-7}}{\varepsilon_0} \mathrm{Vm}^{-1}$$

$$\frac{1 \times 10^{-7}}{\varepsilon_0} \mathrm{Vm}^{-1}$$

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