MHT CET 2023 14th May Morning Shift | Electrostatics Question 3 | Physics

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)

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

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

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

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

MHT CET 2023 13th May Evening Shift

MCQ (Single Correct Answer)

-0

The work done in rotating a dipole placed parallel to the electric field through $$180^{\circ}$$ is W. So, the work done in rotating it through $$60^{\circ}$$ is $$\left(\cos 0^{\circ}=1, \cos 60^{\circ}=\frac{1}{2}, \cos 180^{\circ}=-1\right)$$

$$4 W$$

$$3 \mathrm{~W}$$

$$W / 2$$

$$W / 4$$

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