A positive charge Q is distributed uniformly | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The electric field $E$ on the axis of a ring of radius $R$ at a distance $x$ from the centre is given by $E = \frac{1}{4 \pi \varepsilon_{0}} \fra

Vedclass · Accepted Answer

$\left[\frac{16 \pi^{3} \varepsilon_{0} R^{3} m}{Q q}\right]^{1 / 2}$

Vedclass · Answer

(A) The electric field $E$ on the axis of a ring of radius $R$ at a distance $x$ from the centre is given by $E = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q x}{(R^{2} + x^{2})^{3/2}}$.The force on the particle of charge $-q$ is $F = -qE = -\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q x}{(R^{2} + x^{2})^{3/2}}$.Given $x \ll R$,we can approximate $R^{2} + x^{2} \approx R^{2}$. Thus,$F \approx -\left( \frac{Q q}{4 \pi \varepsilon_{0} R^{3}} \right) x$.This is the equation of simple harmonic motion $(F = -kx)$,where the effective spring constant $k = \frac{Q q}{4 \pi \varepsilon_{0} R^{3}}$.The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{Q q}{4 \pi \varepsilon_{0} R^{3} m}}$.The time period $T$ is $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{4 \pi \varepsilon_{0} R^{3} m}{Q q}} = \sqrt{\frac{4 \pi^{2} \cdot 4 \pi \varepsilon_{0} R^{3} m}{Q q}} = \sqrt{\frac{16 \pi^{3} \varepsilon_{0} R^{3} m}{Q q}}$.

Similar Questions

If ${E_a}$ is the electric field strength of a short dipole at a point on its axial line and ${E_e}$ is the electric field strength on the equatorial line at the same distance,then:

The torque acting on a dipole of moment $\overrightarrow{P}$ in an electric field $\overrightarrow{E}$ is:

The electric field $\vec{E}$ at a point at a distance $r$ from the axis of a dipole is given by:

Three identical dipoles are arranged as shown below. What will be the net electric field at $P$ $\left( {k = \frac{1}{{4\pi {\varepsilon _0}}}} \right)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module