JEE Main 2019 (Online) 12th April Morning Slot | Electrostatics Question 148 | Physics

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-1

A point dipole $$\overrightarrow p = - {p_0}\widehat x$$ is kept at the origin. The potential and electric field due to this dipole on the y-axis at a distance d are, respectively: (Take V= 0 at infinity)

$${{\left| {\overrightarrow p } \right|} \over {4\pi { \in _0}{d^2}}},{{ - \overrightarrow p } \over {4\pi { \in _0}{d^3}}}$$

$$0,{{\overrightarrow p } \over {4\pi { \in _0}{d^3}}}$$

$${{\left| {\overrightarrow p } \right|} \over {4\pi { \in _0}{d^2}}},{{\overrightarrow p } \over {4\pi { \in _0}{d^3}}}$$

$$0,{{ - \overrightarrow p } \over {4\pi { \in _0}{d^3}}}$$

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-1

Shown in the figure is a shell made of a conductor. It has inner radius a and outer radius b, and carries charge Q. At its centre is a dipole $$\overrightarrow P $$ as shown. In this case : JEE Main 2019 (Online) 12th April Morning Slot Physics - Electrostatics Question 149 English

JEE Main 2019 (Online) 12th April Morning Slot Physics - Electrostatics Question 149 English

surface charge density on the inner surface is uniform and equal to $${{\left( {Q/2} \right)} \over {4\pi {a^2}}}$$

surface charge density on the inner surface of the shell is zero everywhere

surface charge density on the outer surface depends on $$\left| {\overrightarrow P } \right|$$

electric field outside the shell is the same as that of a point charge at the centre of the shell

JEE Main 2019 (Online) 10th April Evening Slot

MCQ (Single Correct Answer)

-1

In free space, a particle A of charge 1$$\mu $$C is held fixed at a point P. Another particle B of the same charge and mass 4$$\mu $$g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is : $$\left[ {Take\,{1 \over {4\pi { \in _0}}} = 9 \times {{10}^9}N{m^2}{C^{ - 2}}} \right]$$

1.0 m/s

6.32 $$ \times $$ 10⁴ m/s

2.0 $$ \times $$ 10³ m/s

1.5 $$ \times $$ 10² m/s

JEE Main 2019 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centred at origin. A point charge q is moving towards the ring along the z-axis and has speed u at z = 4a. The minimum value of u such that it crosses the origin is :

$$\sqrt {{2 \over m}} {\left( {{2 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$$

$$\sqrt {{2 \over m}} {\left( {{1 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$$

$$\sqrt {{2 \over m}} {\left( {{1 \over {5}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$$

$$\sqrt {{2 \over m}} {\left( {{4 \over {15}}{{{q^2}} \over {4\pi {\varepsilon _0}a}}} \right)^{1/2}}$$

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