Electrostatics

$\begin{aligned} & F_1=\frac{K Q^2}{r^2}=16 \mathrm{~N} \\ & F_2=\frac{K\left(\frac{Q}{2}\right)\left(\frac{3}{4}\right)}{r^2}=\frac{3}{8} \times 16=6 \mathrm{~N} \end{aligned}$

Final charges on spheres are $\frac{Q}{2}$ and $\frac{3 Q}{4}$.

The question is about calculating the electric field at the surface of a thin spherical shell with a uniform surface charge density denoted by $\sigma$. To determine the electric field at any point on the surface of the shell, we can use Gauss's law, which is particularly useful for systems with high symmetry like a spherical shell.

Gauss's law in its integral form states that the electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space ($ \epsilon_0 $), mathematically represented as:

$ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} $

In the case of a spherical shell of radius $R$, the total charge $Q_{\text{enc}}$ on the shell can be written in terms of the surface charge density $\sigma$ as:

$ Q_{\text{enc}} = \sigma \times 4\pi R^2 $

Now, we apply Gauss's law using a Gaussian surface that coincides with the surface of the spherical shell. The electric field $E$ at the surface of the shell is uniform over the Gaussian surface, and the area of the Gaussian surface (which is also the area of the spherical shell) is $4\pi R^2$. Thus, the electric flux $\Phi_E$ through the Gaussian surface is:

$ \Phi_E = E \cdot 4\pi R^2 $

Using Gauss's law:

$ E \cdot 4\pi R^2 = \frac{\sigma \cdot 4\pi R^2}{\epsilon_0} $

Simplifying this equation gives us the electric field at the surface of the spherical shell:

$ E = \frac{\sigma}{\epsilon_0} $

Therefore, the correct answer is Option B:

$ \frac{\sigma}{\epsilon_0} $

The correct option is Option D: To conduct excess charge due to air friction to ground and prevent sparking.

This method is grounded in the principles of physics, particularly relating to static electricity and grounding. As vehicles move through the air, especially at high speeds, friction between the air and the vehicle can lead to the accumulation of static electricity on the vehicle. This is a common phenomenon and is more pronounced in dry conditions where humidity is low, as moisture in the air can help dissipate static charge more effectively.

In the case of vehicles carrying inflammable fluids, the presence of static electricity poses a significant risk. This is because a static discharge (or sparking) in the presence of flammable vapors can ignite those vapors, leading to potentially catastrophic fires or explosions. The metallic chains that touch the ground serve a crucial safety function by providing a path for the static electrical charge to safely dissipate into the earth, a process known as grounding. The earth acts as a vast reservoir that can absorb large amounts of electrical charge. By grounding the vehicle in this way, the risk of static discharge into the surrounding atmosphere is significantly reduced, enhancing safety by preventing ignition of inflammable materials.

The effectiveness of this safety measure hinges on the electrical conductivity of the chain and its contact with the ground. The chain must be made of a material with good electrical conductivity (such as metal) and maintain adequate contact with the ground to ensure that the static charge can be continuously dissipated. This safety precaution is a practical application of electrostatic principles, showcasing how understanding and harnessing the laws of physics can provide effective solutions to real-world problems.

To find the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system, we use the formulae for both forces and then divide them.

The Coulombic (electrostatic) force, $F_C$, between two charges is given by Coulomb's law:

$F_C = k \frac{|q_1 q_2|}{r^2}$

where $k$ is Coulomb's constant ($8.987 \times 10^9 \, \text{Nm}^2\text{C}^{-2}$), $q_1$ and $q_2$ are the magnitudes of the charges, and $r$ is the distance between the charges. For a proton and an electron, $q_1 = q_2 = e$, where $e$ is the elementary charge ($1.602 \times 10^{-19} \, \text{C}$).

The gravitational force, $F_G$, between two masses is given by Newton's law of universal gravitation:

$F_G = G \frac{m_1 m_2}{r^2}$

where $G$ is the gravitational constant ($6.674 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2}$), and $m_1$ and $m_2$ are the masses of the two objects. For a proton and an electron, $m_p \approx 1.673 \times 10^{-27} \, \text{kg}$ and $m_e \approx 9.109 \times 10^{-31} \, \text{kg}$, respectively.

The ratio of the Coulombic to the gravitational force is therefore:

$\frac{F_C}{F_G} = \frac{k \frac{|e^2|}{r^2}}{G \frac{m_p m_e}{r^2}} = \frac{k e^2}{G m_p m_e}$

Plugging in the values:

$\frac{F_C}{F_G} = \frac{(8.987 \times 10^9) (1.602 \times 10^{-19})^2}{(6.674 \times 10^{-11}) (1.673 \times 10^{-27}) (9.109 \times 10^{-31})}$

$\frac{F_C}{F_G} = \frac{(8.987 \times 10^9) \times (2.568 \times 10^{-38})}{(6.674 \times 10^{-11}) \times (1.523 \times 10^{-57})}$

$\frac{F_C}{F_G} \approx 2.3 \times 10^{39}$

Therefore, the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system is on the order of $10^{39}$, making Option B the correct answer.

When considering the electric flux linked with a cube due to a charge placed at one of its surfaces, it's important to apply Gauss's law. Gauss's law states that the total electric flux through a closed surface is equal to $\frac{q_{\text{enc}}}{\epsilon_0}$, where $q_{\text{enc}}$ is the charge enclosed by the surface and $\epsilon_0$ is the permittivity of free space.

In the given scenario, the charge $q$ is placed at the center of one of the surfaces of the cube. Conceptually, we can think of this arrangement as part of a larger situation where if we had a larger, imaginary cube that encompasses the entire setup such that the point charge is at its geometric center, the charge would then be uniformly distributing its flux through all six faces of this larger cube. However, since the actual setup involves only one cube with one face adjacent to the charge, we essentially have only one-half of the total possible geometry through which the flux from this charge can emerge - implying the flux through our actual cube is a fraction of the total flux that would emanate from the charge if it were centrally placed within a larger, encompassing cube.

Therefore, only half of the flux emanating from the charge will pass through the actual cube because the charge is placed directly on one of its surfaces, effectively distributing its influence through 180 degrees (half of the space around it) instead of the full 360 degrees. Hence, the flux linked with the cube will be half of the total flux $\frac{q}{\epsilon_0}$, which is calculated using Gauss's law for a point charge.

Therefore, the correct answer is $\frac{q}{2\epsilon_0}$.

Option A $\frac{q}{2 \epsilon_0}$ is the correct choice.

The electron revolves in a circle, so its kinetic energy remains same.

So option (2) best represent the given situation.

The electric flux ($\Phi$) through a closed surface, according to Gauss's law, is proportional to the enclosed charge $Q$, and is given by $\Phi = \frac{Q}{\epsilon_0}$, where $\epsilon_0$ is the vacuum permittivity. This is regardless of the shape of the surface, as long as it is closed and it encloses the charge fully.

In this case, both cubes $C_1$ and $C_2$ are closed surfaces, and they are concentric, meaning they share the same center. Each cube encloses a different charge, $2Q$ for $C_1$ and $3Q$ + $2Q$ = $5Q$ for $C_2$. According to Gauss's law, the electric flux through each cube is directly proportional to the enclosed charge.

For $C_1$, enclosing charge $2Q$: $\Phi_{C_1} = \frac{2Q}{\epsilon_0}$

For $C_2$, enclosing charge $5Q$: $\Phi_{C_2} = \frac{5Q}{\epsilon_0}$

Now, we want the ratio of electric flux through $C_1$ to that through $C_2$:

$\frac{\Phi_{C_1}}{\Phi_{C_2}} = \frac{\frac{2Q}{\epsilon_0}}{\frac{5Q}{\epsilon_0}} = \frac{2Q}{5Q} = \frac{2}{5}$

Therefore, the ratio of electric flux passing through $C_1$ to that passing through $C_2$ is $\frac{2}{5}$, which corresponds to option C.

In air $F=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}_2}$

In medium $\mathrm{F}^{\prime}=\frac{1}{4 \pi\left(\mathrm{K} \epsilon_0\right)} \frac{\mathrm{q}_1 \mathrm{q}_2}{\left(\mathrm{r}^{\prime}\right)^2}=\frac{25}{4 \pi\left(5 \epsilon_0\right)} \frac{\mathrm{q}_1 \mathrm{q}_2}{(\mathrm{r})^2}=5 \mathrm{~F}$

$\begin{aligned} & \left(\vec{E}_{\text {net }}\right)_P=0 \\ & \frac{\mathrm{kq}}{\mathrm{x}^2}=\frac{\mathrm{k} \cdot 3 \mathrm{q}}{(\mathrm{r}-\mathrm{x})^2} \\ & (\mathrm{r}-\mathrm{x})^2=3 \mathrm{x}^2 \\ & \mathrm{r}-\mathrm{x}=\sqrt{3} \mathrm{x} \\ & \mathrm{x}=\frac{\mathrm{r}}{\sqrt{3}+1} \end{aligned}$

$\begin{aligned} & \frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{r}}=\mathrm{m} \omega^2 \mathrm{r} \\ & \omega^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\ & \left(\frac{2 \pi}{\mathrm{T}}\right)^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\ & \mathrm{~T}=2 \pi \mathrm{r} \sqrt{\frac{\mathrm{m}}{2 \mathrm{k} \lambda \mathrm{q}}} \end{aligned}$

$V=\frac{k P \cos \theta}{r^2}$

& can also checked dimensionally

$\begin{aligned} & \overrightarrow{\mathrm{E}}=6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{A}}=30 \hat{\mathrm{i}} \\ & \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} \\ & \phi=(6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(30 \hat{\mathrm{i}}) \\ & \phi=6 \times 30=180 \end{aligned}$

The electric flux through any closed surface is given by Gauss's law, which can be stated as:

where:

• $\Phi$ = electric flux
• $Q_{\text{enc}}$ = total charge enclosed by the surface
• $\varepsilon_0$ = permittivity of free space (electric constant)

In this scenario, we have a sphere of radius $4a$ with its center at the origin and two charges, $5Q$ at point (3a, 0) and $-2Q$ at point (-5a, 0). Since the sphere's radius is $4a$, the charge $5Q$, which is located at (3a, 0), lies inside the sphere, whereas the charge $-2Q$, located at (-5a, 0), lies outside the sphere. Gauss's law only considers charges that are enclosed within the surface.

Thus, the charge $-2Q$ has no impact on the electric flux through the sphere because it is not enclosed by the sphere. Only the charge $5Q$ contributes to the electric flux inside the sphere.

The electric flux through the sphere is therefore given simply by the enclosed charge:

The most appropriate answer from the options given would be Option A: Both (A) and (R) are correct, and (R) is the correct explanation of (A).

Here is the reasoning for this answer:

Assertion (A) states that the work done by an electric field on moving a positive charge on an equipotential surface is always zero. This statement is true because by definition, an equipotential surface is a surface over which the electric potential is constant. When a charge moves along an equipotential surface, there is no change in its electric potential energy since potential difference $ \Delta V $ is zero. Work done ($ W $) is defined as the product of charge ($ q $), potential difference ($ \Delta V $), and the cosine of the angle between the field and direction of motion ($ \cos \theta $), which can be written as:

$ W = q \Delta V \cos \theta $

Because $ \Delta V = 0 $ on an equipotential surface, irrespective of the value of $ \cos \theta $, the work $ W $ will be zero. Hence, the Assertion (A) is correct.

Reason (R) says that electric lines of forces are always perpendicular to equipotential surfaces. This statement is also correct as the electric field lines, by definition, are directed such that they are tangent to the electric field vector at any point in space. Since the electric potential is constant on an equipotential surface, there can be no component of the electric field parallel to the surface, as that would imply a force and potential change along the surface. The electric field thus must be perpendicular to the equipotential surface, which means that the electric field lines must also be perpendicular to the equipotential surface. In other words, the electric field does no work when a charge moves along an equipotential surface because the motion is perpendicular to the force.

Therefore, Reason (R) is not only correct, but it is also the correct explanation for Assertion (A), making Option A the right choice.

Potential difference $=\frac{K Q}{r_1}-\frac{K Q}{r_2}$

$\begin{aligned} & r_1=\sqrt{(\sqrt{3})^2+(\sqrt{3})^2} \\ & r_2=\sqrt{(\sqrt{6})^2+0} \end{aligned}$

As $r_1=r_2=\sqrt{6} \mathrm{~m}$

So potential difference $=0$

Consider a circular hoop of radius $a$ with centre $O$.

Bead $A$ is fixed to the hoop, while bead $B$ can slide freely along its circumference.

Forces on bead $B$:

Electrostatic force due to bead $A$, acting along the line $AB$.

Normal reaction from the hoop, directed radially inward toward $O$.

Because these two forces are the only interactions, bead $B$ reaches equilibrium at the point $C$—the position diametrically opposite $A$—where the Coulomb repulsion is exactly balanced by the hoop’s normal reaction.

Displace bead $B$ slightly from its equilibrium point $C$ by a small angle $\theta$ measured at the centre $O$.

1. Geometry of the configuration

In the isosceles triangle $OAB$ (see figure)

$\angle A=\dfrac{\theta}{2}$

$\angle O = 180^{\circ}-\theta$

The side opposite $\angle O$ is $AB$. Using the cosine rule,

$ AB^{2}=a^{2}+a^{2}-2a^{2}\cos(180^{\circ}-\theta)=4a^{2}\cos^{2}\!\frac{\theta}{2}. $

Hence the separation of the charges is

$ r=AB=2a\cos\frac{\theta}{2}. $

2. Coulomb force on $B$

$ F=\frac{q^{2}}{4\pi\varepsilon_{0}r^{2}} =\frac{q^{2}}{16\pi\varepsilon_{0}a^{2}\cos^{2}\dfrac{\theta}{2}}. $

3. Components along the hoop

The force makes an angle $\dfrac{\theta}{2}$ with the radial line $OB$.

Normal component: $F\cos\dfrac{\theta}{2}$  (balanced by the normal reaction of the hoop).

Tangential component: $F\sin\dfrac{\theta}{2}$  (provides the restoring acceleration).

Applying Newton’s second law tangentially,

$ ma\frac{d^{2}\theta}{dt^{2}} =-F\sin\frac{\theta}{2} =-\frac{q^{2}}{16\pi\varepsilon_{0}a^{2}\cos^{2}\dfrac{\theta}{2}}\, \sin\frac{\theta}{2}. $

(The minus sign shows the couple is restoring.)

4. Small-angle approximation

$ \sin\frac{\theta}{2}\simeq\frac{\theta}{2},\qquad \cos\frac{\theta}{2}\simeq1. $

Substituting,

$ \frac{d^{2}\theta}{dt^{2}} =-\frac{q^{2}}{32\pi\varepsilon_{0}ma^{3}}\;\theta =-\omega^{2}\theta, $

where

$ \boxed{\displaystyle \omega^{2} =\frac{q^{2}}{32\pi\varepsilon_{0}ma^{3}} }. $

The motion is simple harmonic, with angular frequency $\omega$ given above.

The electric field inside sphere is zero. So dipole will oscillate when $r>R$.

$\begin{aligned} & \text { For } \mathrm{r}>\mathrm{RE}=\frac{\sigma \mathrm{R}^2}{\varepsilon_0 \mathrm{r}^2} \\ & \omega=\sqrt{\frac{\mathrm{PE}}{\mathrm{I}}}=\sqrt{\frac{\mathrm{P}_0 \sigma \mathrm{R}^2}{\mathrm{I} \varepsilon_0 \mathrm{r}^2}} \end{aligned}$

when $r=2 R$

$\omega=\sqrt{\frac{\mathrm{P}_0 \sigma}{4 \mathrm{I} \varepsilon_0}}$

when $r=10 R$

$\omega=\sqrt{\frac{\mathrm{P}_0 \sigma}{100 \mathrm{I} \varepsilon_0}}$

Given,

$ \begin{array}{l} \mathbf{E}=(8 \hat{\mathbf{i}}+13 \hat{\mathbf{j}}) \mathrm{N} / \mathrm{C} \\\\ A=3 \mathrm{~m}^{2} \\\\ \phi=? \end{array} $

$ \therefore $ Area is lying in the $ X Z $-plane, so vector component of area,

$ \mathbf{A}=3 \hat{\mathbf{j}} $

Now, $\phi=\mathbf{E} \cdot \mathbf{A}$

$ \begin{aligned} & \phi=(8 \hat{\mathbf{i}}+13 \hat{\mathbf{j}}) \cdot(3 \hat{\mathbf{j}}) \\\\ & \phi=(0+39) \mathrm{Wb} \\\\ & \phi=39 \mathrm{~Wb} \end{aligned} $

The work done by an electric field on a charged particle is determined by the equation $ W = F \cdot s $, where $ F $ is the force exerted by the electric field and $ s $ is the displacement of the particle.

The force $ F $ can be expressed as $ F = qE $, where $ q $ is the charge of the particle and $ E $ is the electric field strength. The displacement $ s $ can be determined using the kinematic equation $ s = \frac{1}{2} a t^2 $, where $ a $ is the acceleration and $ t $ is the time.

Substituting these into the work formula gives:

$ W = E \cdot q \cdot \frac{1}{2} a t^2 $

Since both the proton and the alpha particle are in the same electric field and experienced for the same time:

$ W \propto q \cdot a $

Let's denote:

$ W_1 $ as the work on the proton,

$ W_2 $ as the work on the alpha particle.

Hence,

$ \frac{W_1}{W_2} = \frac{q_1 \cdot a_1}{q_2 \cdot a_2} $

For the proton:

$ q_1 = 1 \, \text{C} $,

The acceleration $ a_1 = \frac{E \cdot q_1}{m} $ (where $ m $ is the mass of the proton).

For the alpha particle:

$ q_2 = 2 \, \text{C} $,

The acceleration $ a_2 = \frac{E \cdot (2q)}{4m} = \frac{E}{2} $.

Substituting these into the work ratio equation yields:

$ \frac{W_1}{W_2} = \frac{1 \cdot \frac{E}{E}}{2 \cdot \frac{E}{2E}} $

Simplifying,

$ \frac{W_1}{W_2} = \frac{1}{2} \times 2 = 1:1 $

Thus, the ratio of the work done by the electric field on the proton compared to the alpha particle is $ 1:1 $.

$ Q_{1}=-10 \mu \mathrm{C}, Q_{2}=+5 \mu \mathrm{C} $

$x=0, x=\sqrt{2} \mathrm{~m}$

Net electric field is zero at point $ P $

$ \begin{array}{c} E=\frac{k\left[Q_{1}\right]}{x^{2}}+\frac{k\left[Q_{2}\right]}{(x-\sqrt{2})^{2}}=0 \\\\ \frac{(-10 \mu \mathrm{C}) k}{x^{2}}+\frac{k(5 \mu \mathrm{C}}{(x-\sqrt{2})^{2}}=0 \\\\ \frac{+2}{x^{2}}=\frac{1}{(x-\sqrt{2})^{2}} \end{array} $

Square root both side

$ \begin{aligned} \frac{\sqrt{2}}{x} & =\frac{1}{(x-\sqrt{2})} \\\\ \sqrt{2}(x-\sqrt{2}) & =x \\\\ x(\sqrt{2}-1) & =2 \\\\ x & =\frac{2}{\sqrt{2}-1} \end{aligned} $

Rationalize the denominator

$ \begin{array}{l} x=\frac{2}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} \\\\ x=2(\sqrt{2}+1) \mathrm{m} \end{array} $

From the diagram,

$ \frac{K(8)}{x^2}=\frac{K(32)}{(15+x)^2} \text { (direction of } $

attraction/repulsion is considered)

$ \begin{aligned} & \Rightarrow \quad \frac{1}{x^2}=\frac{4}{(15+x)^2} \\\\ & \Rightarrow \quad \frac{1}{x}=\frac{2}{15+x} \\\\ & \Rightarrow \quad 15+x=2 x \\\\ & \Rightarrow \quad x=15 \mathrm{~cm} \end{aligned} $

From $-8 \mu \mathrm{C}$ charge

Given, Distance of separation, $r=4 \mathrm{~m}$

$ r^2=16 \mathrm{~m} $

Sum of two charges

$ \begin{aligned} q_1+q_2 & =36 \mu \mathrm{C} \\\\ & =36 \times 10^{-6} \mathrm{C} \\\\ F & =0.18 \mathrm{~N} \end{aligned} $

Coulomb's law,

$ F=K \frac{q_1 q_2}{r^2} $

Here, $q_2=36 \times 10^{-6}-q_1$

Thus, $0.18=8.98 \times 10^9 \times \frac{q_1\left(36 \times 10^{-6}-q_1\right)}{16}$

Solving quadratic equation,

$ q_1=\frac{36 \times 10^{-6}+}{\sqrt{\left(36 \times 10^{-6}\right)^2-4 \times \frac{0.18 \times 16}{8.98 \times 10^9}}} $

$\begin{aligned} q_1= & \frac{36 \times 10^{-6}+}{\sqrt{\left(36 \times 10^{-6}\right)^2-4 \times \frac{2.88}{8.98 \times 10^9}}} \\\\ q_1 & =1.98 \times 10^{-5} \mathrm{C} \\\\ q_1 & \approx 20.8 \times 10^{-6} \mathrm{C} \\\\ q_1 & =20 \mu \mathrm{C}\end{aligned}$

Mass, $ m = 0.5 \, \text{g} = 0.5 \times 10^{-3} \, \text{kg} $

Charge, $ q = 10 \times 10^{-6} \, \text{C} $

Electric field, $ E = 8 \, \text{NC}^{-1} $

First, calculate the force $ F $ on the particle using the equation:

$ F = q \times E $

Substituting the values:

$ F = 10 \times 10^{-6} \times 8 = 80 \times 10^{-6} = 8 \times 10^{-5} \, \text{N} $

Next, find the acceleration $ a $ using the formula:

$ a = \frac{F}{m} $

Substitute the values to find:

$ a = \frac{8 \times 10^{-5}}{0.5 \times 10^{-3}} = 0.16 \, \text{m/s}^2 $

Since the particle starts from rest, the initial velocity $ u $ is $ 0 $.

Using the kinematic equation for velocity:

$ v = u + at $

Substituting the known values:

$ v = 0 + 0.16 \times 5 = 0.8 \, \text{m/s} $

Therefore, the velocity of the particle after 5 seconds is $ 0.8 \, \text{m/s} $.

The electric potential $V$ of a sphere is directly proportional to its charge $Q$ and inversely to its radius $r$,

$ V=\frac{k Q}{r} $

For small sphere,

$ \begin{aligned} V_{\text {small }} & =\frac{k Q_{\text {small }}}{r_{\text {small }}}=60 \mathrm{mV} \\ V_{\text {big }} & =\frac{k Q_{\text {total }}}{r_{\text {big }}}=\frac{k \cdot 125 Q_{\text {small }}}{r_{\text {big }}} \end{aligned} $

The volume of the big sphere is 125 times the volume of small sphere, so

$ \begin{aligned} V_{\text {big }} & =\frac{k \times 125 Q_{\text {small }}}{\left(125^{1 / 3}\right) r_{\text {small }}} \\ & =125^{2 / 3} \cdot \frac{k Q_{\text {small }}}{r_{\text {small }}} \\ & =125^{2 / 3} \cdot 60 \mathrm{mV} \\ & =125^{2 / 3} \times 60 \times 10^{-3}=1.5 \mathrm{~V} \end{aligned} $

Therefore, the electric potential on the bigger formed is 1.5 V .

$ \begin{aligned} & \text { Given, } q_1=4 \mathrm{nC}=4 \times 10^{-9} \mathrm{C} \\ & U=1.8 \mu \mathrm{~J}=1.8 \times 10^{-6} \mathrm{~J} \\ & r=10 \mathrm{~cm}=0.1 \mathrm{~m} \end{aligned} $

$ \begin{aligned} & \text { As, } Q=\frac{U \times r}{k \times q_1} \\ & \Rightarrow Q=\frac{1.8 \times 10^{-6} \times 0.1}{8.9875 \times 10^9 \times 4 \times 10^{-9}} \\ & \Rightarrow Q=\frac{1.8 \times 10^{-7}}{35.95} \\ & \Rightarrow Q=5 \times 10^{-9} \mathrm{C} \\ & \Rightarrow Q=5 \mathrm{nC} \end{aligned} $

Given:

Mass of a deuteron, $ m_d = 3.2 \times 10^{-27} \, \text{kg} $

Charge of a deuteron, $ q_d = 1.6 \times 10^{-19} \, \text{C} $

Gravitational acceleration, $ g = 9.8 \, \text{m/s}^2 $

To suspend the deuteron in the air, the electric force ($ F_\text{electric} $) must equal the gravitational force ($ F_\text{gravity} $):

$ F_\text{electric} = F_\text{gravity} $

Thus:

$ qE = mg $

Solving for $ E $, the magnitude of the electric field:

$ E = \frac{mg}{q} = \frac{3.2 \times 10^{-27} \times 9.8}{1.6 \times 10^{-19}} $

$ E = \frac{3.2 \times 9.8}{1.6} \times 10^{-8} $

$ E = 2 \times 9.8 \times 10^{-8} = 19.6 \times 10^{-8} \, \text{N/C} $

Therefore, the magnitude of the electric field required is $ 19.6 \times 10^{-8} \, \text{N/C} $.

Given:

$ q_1 = 5 \, \text{nC} = 5 \times 10^{-9} \, \text{C} $

$ q_2 = -2 \, \text{nC} = -2 \times 10^{-9} \, \text{C} $

These charges are positioned along the X-axis. First, let's determine the distance between them:

$ \begin{align*} x &= x_2 - x_1 \\ &= (23 - 5) \, \text{cm} = 18 \, \text{cm} = 18 \times 10^{-2} \, \text{m} \end{align*} $

Now, calculate the electrostatic potential energy ($U$) of the charges using the formula:

$ U = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{x} $

Substitute the known values:

$ U = \frac{9 \times 10^9 \times 5 \times 10^{-9} \times (-2 \times 10^{-9})}{18 \times 10^{-2}} $

$ U = -5 \times 10^{-7} \, \text{J} $

The negative sign in the electrostatic potential energy indicates that the force between the two charges is attractive.

Given:

Electric dipole moment, $ p = 5 \times 10^{-30} \, \text{C} \cdot \text{m} $

In a neutral ammonia (NH$_3$) molecule, there are 10 electrons and 10 protons. The electric dipole moment is expressed as:

$ p = q \times 2l = 10e \times 2l $

Where $ e $ is the elementary charge ($1.6 \times 10^{-19} \, \text{C}$).

To find the separation distance $ 2l $, we use:

$ 2l = \frac{p}{10e} = \frac{5 \times 10^{-30}}{10 \times 1.6 \times 10^{-19}} $

Calculating this gives:

$ 2l = 3.125 \times 10^{-12} \, \text{m} $

Thus, the distance between the centers of positive and negative charge in the molecule is $ 3.125 \times 10^{-12} \, \text{m} $.

$ \begin{aligned} & \text { } q_1=+1 \times 10^{-8} \mathrm{C} \\ & q_2=-2 \times 10^{-8} \mathrm{C} \\ & q_3=+3 \times 10^{-8} \mathrm{C} \\ & q_4=+2 \times 10^{-8} \mathrm{C} \end{aligned} $

Here,

At point $A$ and $C$, charges are $q_1=1 \times 10^{-8} \mathrm{C}$ and $3 \times 10^{-1} \mathrm{C}$

$ \begin{aligned} V_{A C} & =9 \times 10^9\left[\frac{1 \times 10^{-8}+3 \times 10^{-8}}{1}\right] \sqrt{2} \\ & =9 \times 10^9\left[4 \times 10^{-8}\right] \sqrt{2} \end{aligned} $

$ \begin{aligned} & =9 \times 10^9 \times 4 \times 10^{-8} \times \sqrt{2} \\ & =90 \times 4 \times \sqrt{2}=360 \sqrt{2}=510 \mathrm{~V} \end{aligned} $

At point $B$ and $D$, charges are $q_2=-2 \times 10^{-8} \mathrm{C}$ and $q_4=2 \times 10^{-8} \mathrm{C}$ Thus potential $V_{B D}=0$ Hence, answer will be 510 V

Given:

Distance $ r = 3 \, \text{m} $

Linear charge density $ \lambda = 0.2 \, \mu\text{C/m} = 0.2 \times 10^{-6} \, \text{C/m} $

The electric field $ E $ at a distance $ r $ from a long straight wire is calculated using the formula:

$ E = \frac{\lambda}{2\pi \varepsilon_0 r} $

Where:

$\varepsilon_0$ is the permittivity of free space, approximately $8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2$

Substitute the given values into the formula:

$ E = \frac{0.2 \times 10^{-6}}{2 \pi \times 8.85 \times 10^{-12} \times 3} $

This calculation results in:

$ E \approx 1.2 \times 10^3 \, \text{V/m} $

Given:

A charge $ q $ is placed at the center of a cube.

The side length of the cube is $ L $.

We know that the electric flux through a closed surface that encloses a charge $ q $ is given by Gauss’s Law:

$ \phi_t = \frac{q}{\varepsilon_0} $

Here, $\phi_t$ represents the total electric flux through the entire surface of the cube.

Since a cube has 6 faces, the flux through each individual face, $\phi_f$, is the total flux divided equally among the six faces:

$ \phi_f = \frac{1}{6} \times \phi_t = \frac{1}{6} \times \frac{q}{\varepsilon_0} = \frac{q}{6 \varepsilon_0} $

This shows that the electric flux associated with each face of the cube is $\frac{q}{6 \varepsilon_0}$.

Given:

Each side of the equilateral triangle has a length $L$.

Each vertex has a charge $q$.

The potential energy $U$ of the system can be calculated by considering the interactions between each pair of charges. Since there are three sides in an equilateral triangle, we have three pairs of charges to consider:

$ U = \frac{1}{4 \pi \varepsilon_0}\left[\frac{q \cdot q}{L} + \frac{q \cdot q}{L} + \frac{q \cdot q}{L}\right] $

Simplifying the expression:

$ U = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{L} \cdot (1 + 1 + 1) $

$ U = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{3q^2}{L} $

Therefore, the potential energy of the system of three equal charges at the vertices of an equilateral triangle is:

$ U = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{3q^2}{L} $

To understand the change in capacitance when eight smaller drops combine to form a larger drop, consider the following:

Let $ R $ be the radius of the larger drop and $ r $ be the radius of each smaller drop. When these smaller drops combine, the total volume remains constant. Therefore, the volume of the large drop equals the total volume of the eight smaller drops:

$ \frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi r^3 $

This simplifies to:

$ R^3 = 8r^3 $

$ R = 2r $

Given that the capacitance $ C $ of an isolated spherical conductor is described by the formula:

$ C = 4 \pi \varepsilon_0 r $

The capacitance of each smaller drop is:

$ C = 4 \pi \varepsilon_0 r $

For the larger drop, now having a radius $ R = 2r $, the capacitance $ C' $ is calculated as follows:

$ C' = 4 \pi \varepsilon_0 R $

$ C' = 4 \pi \varepsilon_0 (2r) = 2 \times 4 \pi \varepsilon_0 r = 2C $

Thus, the capacitance of the larger drop is $ 2C $.