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 $=0$

The electric potential (V) at a distance (r) from a point charge (Q) is given by the formula:

$ V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} $

In this case, we know (V) and (Q), and we're asked to solve for (r). We can rearrange the formula to solve for (r):

$ r = \frac{1}{4\pi\epsilon_0} \frac{Q}{V} $

Substituting the given values into this equation gives:

$ r = \frac{9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2}}{1} \frac{5 \times 10^{-9} \, \text{C}}{50 \, \text{V}} = 0.9 \, \text{m} $

So, the distance of the point P from the point charge is 0.9 meters. This corresponds to 90 cm.

${E_I} \ne 0$ (inside region)

${E_{II}} = 0$ (conducting region)

${E_{III}} \ne 0$

$ = {{KQ} \over {{r^2}}}\,\,\,(r > b)$

$\overrightarrow E = \left( {{A \over {{x^2}}}\widehat i + {B \over {{y^3}}}\widehat j} \right)$

$\left[ {{A \over {{x^2}}}} \right] = [E] = \left[ {{F \over q}} \right] = \left[ {{N \over C}} \right] = N{C^{ - 1}}$

$\mathrm{[A] = (N{m^2}{C^{ - 1}})}$

$\mathrm{[B] = N{m^3}{C^{ - 1}}}$

$\sigma = {{{Q_1}} \over {4\pi {R^2}}} = {{{Q_2}} \over {4\pi {{(2R)}^2}}}$

Now $Q{'_2} = {Q_{total}}\left[ {{{{R_2}} \over {{R_1} + {R_2}}}} \right]$

$ = ({Q_1} + {Q_2})\left[ {{{2R} \over {3R}}} \right]$

$ = \sigma (20\pi {R^2}){2 \over 3}$

$\therefore$ $\sigma {'_2} = {{Q{'_2}} \over {4\pi {{(2R)}^2}}} = {{\sigma (20\pi {R^2}){2 \over 3}} \over {16\pi {R^2}}}$

$ = {5 \over 4} \times {2 \over 3}\sigma $

$ = {5 \over 6}\sigma $

$w = \int {F\,.\,ds} $

$\overrightarrow F = q\overrightarrow E = 2 \times {10^{ - 2}} \times 30\widehat i = 0.6N\widehat i$

$w = \overrightarrow F .\overrightarrow d = (0.6\widehat i)\,.\,( - \widehat i, - 2\widehat j)$

$ = - 0.6$ J

$ = - 600$ mJ

Let two identical spheres have charge q. And distance between them = r

$\therefore$ Force between the spheres $(F) = {{k{q^2}} \over {{r^2}}}$

Now when an identical uncharged sphere C comes in contact with A, charge q on sphere A get's divided equally to both sphere.

So, both sphere A and C have charge $ = {q \over 2}$

Now, C get's in contact with B. So their total charge $\left( {q + {q \over 2}} \right)$ gets divided equally.

So, charge on both B and C is $ = {{q + {q \over 2}} \over 2} = {{3q} \over 4}$

Now, C is place midpoint between A and B.

Repulsion force between A and C,

${F_{AC}} = {{k\left( {{q \over 2}} \right)\left( {{{3q} \over 4}} \right)} \over {{{\left( {{r \over 2}} \right)}^2}}} = {{4k \times 3{q^2}} \over {8{r^2}}}$

Repulsion force between B and C,

${F_{BC}} = {{k\left( {{{3q} \over 4}} \right)\left( {{{3q} \over 4}} \right)} \over {{{\left( {{r \over 2}} \right)}^2}}} = {{4k \times 9{q^2}} \over {16{r^2}}}$

$\therefore$ Net force on C,

${F_{net}} = {F_{BC}} - {F_{AC}}$

$ = {{4k \times 9{q^2}} \over {16{r^2}}} - {{4k \times 3{q^2}} \over {8{r^2}}}$

$ = {{k{q^2}} \over {{r^2}}}\left( {{9 \over 4} - {3 \over 2}} \right)$

$ = {{k{q^2}} \over {{r^2}}} \times {3 \over 4}$

$ = {3 \over 4}F$ [as $F = {{k{q^2}} \over {{r^2}}}$]

$\left( {4\pi {r^2}} \right){E_\rho } = {{{Q_{in}}} \over {{\varepsilon _0}}}$

$ = {{\int_0^r {{\rho _0}\left( {{3 \over 4} - {r \over 4}} \right)4\pi {r^2}dr} } \over {{\varepsilon _0}}}$

$ = {{{\rho _0}\pi 4} \over {{\varepsilon _0}}}\left( {{{{r^3}} \over 4} - {{{r^4}} \over {4R}}} \right)$

${E_\rho } = {{{\rho _0}} \over {4{\varepsilon _0}}}\left( {r - {{{r^2}} \over R}} \right)$

$ = {{{\rho _0}r} \over {4{\varepsilon _0}}}\left( {1 - {r \over R}} \right)$

Since ${\overrightarrow E _{net}} = \overrightarrow 0 $ in the bulk of a conductor

$\Rightarrow$ Potential would be constant.

$\Rightarrow$ Statement I is correct

Since a conductor's surface is equipotential, $\overrightarrow E $ just outside is perpendicular to the surface.

$E = {{8\,m} \over e}$ V/m

$l = 1$ m

${v_x} = 2$ m/s

${a_y} = - 8$ m/s²

$t = {l \over {{v_x}}} = {1 \over 2}s$

$ \Rightarrow |{v_y}| = 4$ m/s

$\Rightarrow$ angle of deviation = $\theta$

$\tan \theta = {{{v_y}} \over {{v_x}}}$

$\theta = {\tan ^{ - 1}}\left( {{4 \over 2}} \right) = {\tan ^{ - 1}}(2)$

so $F = {{kq(4 - q) \times {{10}^{ - 12}}} \over {{r^2}}}$

so F_max will be at q = 2 $\mu$C

(x << a) ($\alpha$ is acceleration)

${F_{net}} = - \left( {{{k{q_0}Q} \over {{{(a - x)}^2}}} - {{kQ{q_0}} \over {{{(a + x)}^2}}}} \right)$

$m\alpha = - {{k{q_0}Q} \over {{a^4}}}4ax$

$ \Rightarrow \alpha = - {{4k{q_0}Q} \over {m{a^3}}}x$

So, $T = 2\pi \sqrt {{{4\pi {\varepsilon _0}m{a^3}} \over {4{q_0}Q}}} $

or $T = \sqrt {{{4{\pi ^3}{\varepsilon _0}m{a^3}} \over {{q_0}Q}}} $

After connection

${\sigma _1}{R_1} = {\sigma _2}{R_2}$

Now $E = {\sigma \over {{\varepsilon _0}}}$

$ \Rightarrow {{{E_1}} \over {{E_2}}} = {{{\sigma _1}} \over {{\sigma _2}}} = {{{R_2}} \over {{R_1}}} = {2 \over 1}$

Force experienced by the charge q

$F = {{kQqx} \over {{{\left[ {{{\left( {{d \over 2}} \right)}^2} + {x^2}} \right]}^{{3 \over 2}}}}}$

For maximum Coulomb's force for x

${{dF} \over {dx}} = 0$

On solving $x = {d \over {2\sqrt 2 }}$

$\overrightarrow E = -{{dV} \over {dx}}\widehat i$

$\overrightarrow E = - 6x\widehat i$

So, $\overrightarrow E $ at (1, 0, 3) is

$\overrightarrow E = - 6\widehat i$ V/m

${v^2} - {u^2} = 2as$

$ \Rightarrow {0^2} - {200^2} = 2\left( {{{ - qE} \over m}} \right)(S)$

$ \Rightarrow - {200^2} = 2\left[ {{{ - 40 \times {{10}^{ - 6}} \times {{10}^5}} \over {100 \times {{10}^{ - 6}}}}} \right][S]$

$ \Rightarrow S = {4 \over {2 \times 4}}$ m = 0.5 m

Electric field at P will be

$E = {{kq} \over {{{(d/2)}^2}}} \times 2 = {{8kq} \over {{d^2}}}$

So, ${{8 \times 9 \times {{10}^9} \times 8 \times {{10}^{ - 6}}} \over {{d^2}}} = 6.4 \times {10^4}$

So, $d = 3$ m

As one moves closer to a positive charge (isolated) the density of electric field line increases and so does the electric field intensity

$\Rightarrow$ Statement I is true

As opposite poles of an electric dipole would experience equal and opposite forces so net force on a dipole in a uniform electric field will be zero

$\Rightarrow$ Statement II is true

$\left| {{E_0}} \right| = {{kq/2} \over {{a^2}}}\sqrt 2 + {{kq} \over {{{\left( {a\sqrt 2 } \right)}^2}}}$

$ = {{kq} \over {\sqrt 2 {a^2}}} + {{kq} \over {2{a^2}}}$

$ = {{kq} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }} + {1 \over 2}} \right),\,k = {1 \over {4\pi {\varepsilon _0}}}$

Flux passing through flat surface = Flux passing through curved surface.

So, $\phi = {q \over {2{\varepsilon _0}}}$

F_net on charge 3, ${F_1} = {{\sqrt 3 k{q^2}} \over {{1^2}}}$

Force between any 2 charges

${F_2} = {{k{q^2}} \over {{1^2}}}$

So, ${{{F_1}} \over {{F_2}}} = \sqrt 3 $