Electrostatics · Physics · NEET

An electric dipole with dipole moment $5 \times 10^{-6} \mathrm{Cm}$ is aligned with the direction of a uniform electric field of magnitude $4 \times 10^5 \mathrm{~N} / \mathrm{C}$. The dipole is then rotated through an angle of $60^{\circ}$ with respect to the electric field. The change in the potential energy of the dipole is:

Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is $q$ and the force of repulsion between them is $F$. A third identical uncharged conducting sphere is brought in contact with sphere $A$ first and then with $B$ and finally removed from both. New force of repulsion between spheres $A$ and $B$ (Radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:

A metal cube of side $$5 \mathrm{~cm}$$ is charged with $$6 \mu \mathrm{C}$$. The surface charge density on the cube is

The value of electric potential at a distance of $$9 \mathrm{~cm}$$ from the point charge $$4 \times 10^{-7} \mathrm{C}$$ is [Given $$\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}$$] :

A thin spherical shell is charged by some source. The potential difference between the two points $$C$$ and $$P$$ (in V) shown in the figure is:

(Take $$\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9$$ SI units)

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: The potential (V) at any axial point, at $$2 \mathrm{~m}$$ distance $$(r)$$ from the centre of the dipole of dipole moment vector $$\vec{P}$$ of magnitude, $$4 \times 10^{-6} \mathrm{C} \mathrm{m}$$, is $$\pm 9 \times 10^3 \mathrm{~V}$$.

(Take $$\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}$$ units)

Reason R: $$V= \pm \frac{2 P}{4 \pi \epsilon_0 r^2}$$, where $$r$$ is the distance of any axial point, situated at $$2 \mathrm{~m}$$ from the centre of the dipole.

In the light of the above statements, choose the correct answer from the options given below:

According to Gauss law of electrostatics, electric flux through a closed surface depends on :

A charge $$\mathrm{Q} ~\mu \mathrm{C}$$ is placed at the centre of a cube. The flux coming out from any one of its faces will be (in SI unit) :

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)

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 $$2 \mathrm{~cm}$$.

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

An electric dipole is placed as shown in the figure.

The electric potential (in 10² V) at point P due to the dipole is ($$\in_0$$ = permittivity of free space and $$\frac{1}{4 \pi \epsilon_{0}}$$ = K) :

Six charges +q, $$-$$q, +q, $$-$$q, +q, and $$-$$q are fixed at the corners of a hexagon of side d as shown in the figure. The work done in bringing a charge q₀ to the centre of the hexagon from infinity is

($${\varepsilon _0}$$ - permittivity of free space)

The angle between the electric lines of force and the equipotential surface is

Two hollow conducting spheres of radii R₁ and R₂ (R₁ >> R₂) have equal charges. The potential would be

Two point charges $$-$$q and +q are placed at a distance of L, as shown in the figure.

The magnitude of electric field intensity at a distance R(R >> L) varies as: