Electrostatics · Physics · NEET
MCQ (Single Correct Answer)
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:

$$\left( {{1 \over {4\pi {\varepsilon _0}}} = 9 \times {{10}^9}N{m^2}/{C^2}} \right)$$
$$\left( {{1 \over {4\pi {\varepsilon _0}}} = 9 \times {{10}^9}N{m^2}/{c^2}} \right)$$

A positive charge is moved from A to B in each diagram.
[Given : mass of hydrogen mh = 1.67 $$ \times $$ 10$$-$$27 kg]





The electric flux linked to the surface, in units of volt m is
The electric field at that point is





The change in the potential energy of the system is $${{{q_3}} \over {4\pi {\varepsilon _0}}}$$ where k is

