Electric Charges and Fields

Let the distance between them is r.

According to the conservation of charge, the charge on each after contact is

$$\begin{gathered} \mathrm{q} = \frac{4\times {10}^{-8}}{2}=2\times {10}^{-8} \mathrm{C} \\ \mathrm{Now} \\ \mathrm{F} =\frac{\mathrm{k} {\mathrm{q}}^2}{{\mathrm{r}}^2} \\ 9\times {10}^{-3} = \frac{9\times {10}^{-3}\times 2\times {10}^{-8}\times 2\times {10}^{-8}}{{\mathrm{r}}^2} \\ \mathrm{r} = 0.02 \mathrm{m} \end{gathered}$$

Concept Used-Charge Redistribution

$Q_T$=$Q_1$$+$$Q_2$$+$$Q_3$

$Q_T$= -7+4+$Q_3$$\dots \dots$(1)

after contact

$Q_F$=$\frac{Q_T}{3}$= -2pc

$Q_T$= -6pc $\dots \dots$(2)

-6= -7+4+$Q_3$

$Q_3$= -3pc

The acceleration a of each particle is given by Newton’s second law:F=ma and F=qE ⇒a=qE/m.
Thus, the time taken for each particle to travel the same distance is given by the kinematic equation:

d=$\frac{1}{2}a{t}^2$

⇒t=$\sqrt{\frac{2d}{a}}$=$\sqrt{\frac{2d}{{\displaystyle \frac{qE}{m}}}}$
The ratio of times tp and tq for particles P and Q is:$\sqrt{\frac{m_P}{m_Q}}$

The acceleration a of each particle is given by Newton’s second law:F=ma and F=qE ⇒a=qE/m.
Thus, the time taken for each particle to travel the same distance is given by the kinematic equation:

d=$\frac{1}{2}a{t}^2$

⇒t=$\sqrt{\frac{2d}{a}}$=$\sqrt{\frac{2d}{{\displaystyle \frac{qE}{m}}}}$
The ratio of times tp and tq for particles P and Q is:$\sqrt{\frac{m_P}{m_Q}}$

The acceleration a of each particle is given by Newton’s second law:F=ma and F=qE ⇒a=qE/m.
Thus, the time taken for each particle to travel the same distance is given by the kinematic equation:

d=$\frac{1}{2}a{t}^2$

⇒t=$\sqrt{\frac{2d}{a}}$=$\sqrt{\frac{2d}{{\displaystyle \frac{qE}{m}}}}$
The ratio of times tp and tq for particles P and Q is:$\sqrt{\frac{m_P}{m_Q}}$

Given Data:

Length of the plates:

$L=10cm=0.1m$

Initial horizontal velocity: $v_x={10}^{6}m/s$
Electric field: $E=9.1V/cm=910V/m$
Charge of an electron: $e=1.6\times {10}^{-19}C$
Mass of an electron: $m=9.1\times {10}^{-31}kg$
The force on the electron due to the electric field is given by:

$$\begin{gathered} F=eE=\left(1.6\times {10}^{-19}\right)\times \left(910\right)=1.456\times \\ {10}^{-16}N \\ a=\frac{F}{m}=\frac{1.456\times {10}^{-16}}{9.1\times {10}^{-31}} \\ a=1.6\times {10}^{14}m/s^2 \end{gathered}$$

The time t taken by the electron to travel through the plates is given by:

$t=\frac{L}{v_x}=\frac{0.1}{{10}^6}={10}^{-7}s$

Since the electron is initially moving horizontally, its initial vertical velocity is 0 . The vertical velocity is given by:

$$\begin{gathered} v_y=at \\ {v}_y=\left(1.6\times {10}^{14}\right)\times \left({10}^{-7}\right) \\ {v}_y=16\times {10}^{6}m/s \end{gathered}$$

The electrostatic potential due to an electronic dipole at a distance r given as $V_P\propto \frac{1}{r^2}$

The electrostatic force is given by

${\mathrm{F}}_{\mathrm{E}} = \frac{{\mathrm{q}}_1{\mathrm{q}}_2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}} {\mathrm{r}}^2}$

The gravitational force is given by

${\mathrm{F}}_{\mathrm{G}} = \frac{\mathrm{G} {\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2}$

Ratio of these forces is given by

$\frac{{\mathrm{F}}_{\mathrm{E}}}{{\mathrm{F}}_{\mathrm{G}}}=\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{Gm}}_1{\mathrm{m}}_2}$

A ring of radius R is uniformly charged with linear charge density $\lambda$.
Total charge on the ring: $\mathrm{Q} = \lambda \mathit{ }\left(2\pi R\right)$
By symmetry, the electric field components at the centre due to opposite elements of the ring cancel out.
Hence, the net electric field at the centre of the ring is zero.
E = 0

Concept Used-Charge Redistribution

$Q_T$=$Q_1$$+$$Q_2$$+$$Q_3$

$Q_T$= -7+4+$Q_3$$\dots \dots$(1)

after contact

$Q_F$=$\frac{Q_T}{3}$= -2pc

$Q_T$= -6pc $\dots \dots$(2)

-6= -7+4+$Q_3$

$Q_3$= -3pc

A ring of radius R is uniformly charged with linear charge density $\lambda$.
Total charge on the ring: $\mathrm{Q} = \lambda \mathit{ }\left(2\pi R\right)$
By symmetry, the electric field components at the centre due to opposite elements of the ring cancel out.
Hence, the net electric field at the centre of the ring is zero.
E = 0

Let the distance between them is r.

According to the conservation of charge, the charge on each after contact is

$$\begin{gathered} \mathrm{q} = \frac{4\times {10}^{-8}}{2}=2\times {10}^{-8} \mathrm{C} \\ \mathrm{Now} \\ \mathrm{F} =\frac{\mathrm{k} {\mathrm{q}}^2}{{\mathrm{r}}^2} \\ 9\times {10}^{-3} = \frac{9\times {10}^{-3}\times 2\times {10}^{-8}\times 2\times {10}^{-8}}{{\mathrm{r}}^2} \\ \mathrm{r} = 0.02 \mathrm{m} \end{gathered}$$

$$\begin{gathered} \ddot{u}{\omega }^2\theta =\ell {}_0\theta \\ 2m{\left(\frac{\ell }{2}\right)}^2{\omega }^2=q\ell E_0 \\ {\omega }^2=\frac{2q{E}_0}{m\ell } \\ T=2\pi \sqrt{\frac{m\ell }{2q{E}_0}} \end{gathered}$$

In metal all the charge is present on surface.

\(\mathrm{Q}=6 \mu C\)

Total surface area \(S=6 a^2\)

\(\begin{aligned}
& =6 \times\left(5 \times 10^{-2}\right)^2 \\
& =6 \times 25 \times 10^{-4} \\
& =150 \times 10^{-4} \mathrm{~m}^2
\end{aligned}\)

Surface charge density \(\sigma=\frac{Q}{S}\)

\(\begin{aligned}
& =\frac{6 \times 10^{-6}}{150 \times 10^{-4}} \\
& =0.4 \times 10^{-3} \mathrm{Cm}^{-2}
\end{aligned}\)

$$\begin{gathered} \ddot{u}{\omega }^2\theta =\ell {}_0\theta \\ 2m{\left(\frac{\ell }{2}\right)}^2{\omega }^2=q\ell E_0 \\ {\omega }^2=\frac{2q{E}_0}{m\ell } \\ T=2\pi \sqrt{\frac{m\ell }{2q{E}_0}} \end{gathered}$$

A ring of radius R is uniformly charged with linear charge density $\lambda$.
Total charge on the ring: $\mathrm{Q} = \lambda \mathit{ }\left(2\pi R\right)$
By symmetry, the electric field components at the centre due to opposite elements of the ring cancel out.
Hence, the net electric field at the centre of the ring is zero.
E = 0

The electric force is given by:

$$F_E = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}.$$

The gravitational force is given by:

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

The ratio of these forces is:

$$\frac{F_E}{F_G} = \frac{\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}}{G \frac{m_1 m_2}{r^2}} = \frac{q_1 q_2}{4 \pi \epsilon_0 G m_1 m_2}.$$​​.

Thus, the correct answer is (d).

Concept Used-Charge Redistribution

$Q_T$=$Q_1$$+$$Q_2$$+$$Q_3$

$Q_T$= -7+4+$Q_3$$\dots \dots$(1)

after contact

$Q_F$=$\frac{Q_T}{3}$= -2pc

$Q_T$= -6pc $\dots \dots$(2)

-6= -7+4+$Q_3$

$Q_3$= -3pc

The electric force is given by:

$$F_E = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}.$$

The gravitational force is given by:

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

The ratio of these forces is:

$$\frac{F_E}{F_G} = \frac{\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}}{G \frac{m_1 m_2}{r^2}} = \frac{q_1 q_2}{4 \pi \epsilon_0 G m_1 m_2}.$$​​.

Thus, the correct answer is (d).

According to Gauss's Law,

$$\begin{gathered} \Phi =\frac{q}{{ϵ}_0} \\ \text{ or, }q=\Phi \times {ϵ}_0 \\ q=\left(-2\times {10}^4\right)\times \left(8.85\times {10}^{-12}\right) \\ q=-17.7\times {10}^{-8}C \end{gathered}$$

The acceleration a of each particle is given by Newton’s second law:F=ma and F=qE ⇒a=qE/m.
Thus, the time taken for each particle to travel the same distance is given by the kinematic equation:

d=$\frac{1}{2}a{t}^2$

⇒t=$\sqrt{\frac{2d}{a}}$=$\sqrt{\frac{2d}{{\displaystyle \frac{qE}{m}}}}$
The ratio of times tp and tq for particles P and Q is:$\sqrt{\frac{m_P}{m_Q}}$

The electric flux \(\phi\) has dimensions of charge per unit area, i.e.,:

$\left[\varphi \right]=\left[\frac{Q}{\text{ Area }}\right]$

a is associated with surface charge density s , which has the dimensions:

$\left[\sigma \right]=\frac{Q}{\text{ Area }}$

Thus, the dimensions of a will be:

$\left[\alpha \right]=\left[\frac{\varphi }{\sigma }\right]=\frac{Q/\text{ Area }}{Q/\text{ Area }}=1$

Next, for 1 , the linear charge density, we have:

$\left[\lambda \right]=\frac{Q}{\text{ Length }}$

Since b is associated with 1 , the dimensions of b are:

$\left[\beta \right]=\left[\frac{\varphi }{\lambda }\right]=\frac{Q/\text{ Area }}{Q/\text{ Length }}=\frac{\text{ Length }}{\text{ Area }}$

Now, considering the ratio $\frac{\alpha }{\beta }$, we find:

$\left[\frac{\alpha }{\beta }\right]=\frac{1}{\frac{\text{ Length }}{\text{ Area }}}=\text{ Length }$

Thus, the ratio $\frac{\alpha }{\beta }$ represents length.

• A charged particle of charge Q and mass $m$m is suspended from a string of length $l$.
• It is placed in a uniform electric field $E$​.
• Gravity is ignored.
• The particle will undergo small oscillations due to the torque exerted by the electric field.

• The electric force acting on the charge is:

  $F=QE$​

• The torque about the hinge when displaced by an angle $\theta$ is:

  $\tau =QE\cdot l\sin \theta$

• For small angles, using $\sin \theta \approx \theta$

  $\tau \approx QEl\theta$

• Comparing with the standard equation for simple harmonic motion:

  $$\tau = - k \theta$$

  The restoring torque constant is:

  $k=QEl$

• The equation for angular motion:

  $$I \alpha = - k \theta$$I

  where:

  • $I = ml^2$ is the moment of inertia of the particle about the hinge.
  • $\alpha$ is the angular acceleration.
  $m{l}^2\frac{d^2\theta }{d{t}^2}=-QEl\theta$
• $\frac{d^2\theta }{d{t}^2}+\frac{QE}{ml}\theta =0$

  This is the standard form of SHM with angular frequency $\omega$:

  $\omega =\sqrt{\frac{QE}{ml}}$​

The period is given by:

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{ml}{QE}}$​

The correct option is:

$$\boxed{A} \quad T = 2\pi \sqrt{\frac{m l}{Q E}}$$​

• A charged particle of charge Q and mass $m$m is suspended from a string of length $l$.
• It is placed in a uniform electric field $E$​.
• Gravity is ignored.
• The particle will undergo small oscillations due to the torque exerted by the electric field.

• The electric force acting on the charge is:

  $F=QE$​

• The torque about the hinge when displaced by an angle $\theta$ is:

  $\tau =QE\cdot l\sin \theta$

• For small angles, using $\sin \theta \approx \theta$

  $\tau \approx QEl\theta$

• Comparing with the standard equation for simple harmonic motion:

  $$\tau = - k \theta$$

  The restoring torque constant is:

  $k=QEl$

• The equation for angular motion:

  $$I \alpha = - k \theta$$I

  where:

  • $I = ml^2$ is the moment of inertia of the particle about the hinge.
  • $\alpha$ is the angular acceleration.
  $m{l}^2\frac{d^2\theta }{d{t}^2}=-QEl\theta$
• $\frac{d^2\theta }{d{t}^2}+\frac{QE}{ml}\theta =0$

  This is the standard form of SHM with angular frequency $\omega$:

  $\omega =\sqrt{\frac{QE}{ml}}$​

The period is given by:

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{ml}{QE}}$​

The correct option is:

$$\boxed{A} \quad T = 2\pi \sqrt{\frac{m l}{Q E}}$$​