MHT CET 2021 21th September Evening Shift | Electrostatics Question 44 | Physics

A hollow metal sphere has a radius 'r'. The potential difference between a point on its surface and at a point at a distance '3r' from its centre is 'V'. The electric intensity at the distance '3r' from the centre of the sphere will be :

$$\mathrm{\frac{V}{3r}}$$

3Vr

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

$$\mathrm{\frac{V}{6r}}$$

MHT CET 2021 21th September Morning Shift

MCQ (Single Correct Answer)

-0

'$$\mathrm{F}$$' is the force between the two identical charged particles placed at a distance '$$\mathrm{Y}$$' from each other. If the distance between the charges is reduced to half the previous distance then force between them becomes

$$\frac{F}{4}$$

$$4 \mathrm{~F}$$

$$2 \mathrm{~F}$$

$$\frac{\mathrm{F}}{2}$$

MHT CET 2021 20th September Evening Shift

MCQ (Single Correct Answer)

-0

A uniformly charged semicircular arc of radius '$$r$$' has linear charge density $$(\lambda)$$, is the electric field at its centre? ( $$\in_0=$$ permittivity of free space)

$$\frac{\lambda}{4 \epsilon_0 r}$$

$$\frac{2 \pi \epsilon_0}{\lambda}$$

$$\frac{\lambda}{4 \epsilon_0}$$

$$\frac{2 \epsilon_0}{\lambda}$$

MHT CET 2021 20th September Evening Shift

MCQ (Single Correct Answer)

-0

A hollow charged metal sphere has radius 'R'. If the potential difference between its surface and a point at a distance '5 R' from the centre is $$\mathrm{V}$$, then magnitude of electric field Intensity at a distance '5R' from the centre of sphere is

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

$$\frac{\mathrm{V}}{20 \mathrm{R}}$$

$$10 \mathrm{VR}$$

$$20 \mathrm{VR}$$

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