Question

Consider a uniformly charged ring in the xy plane, centered at the origin. The ring has radius a and positive charge q distributed evenly along its circumference. 

PartA

What is the direction of the electric fieldat any point on the z axis?

	parallel to the x axis
	parallel to the y axis
	parallel to the z axis
	in a circle parallel to the xy plane

PartB

What is the magnitude of the electric fieldalong the positive z axis?

Use k in your answer, where $k = \frac{1}{4 \pi \epsilon_0}$ .

E(z) =

PartC

Imagine a small metal ball of mass m and negative charge -q₀. The ball is released from rest at the point(0,0,d) and constrained to movealong thez axis, with no damping. If  0<d≤a, what will be theball's subsequent trajectory?

	repelled from the origin
	attracted toward the origin and coming to rest
	oscillating along the z axis between z=d and z=-d
	circling around the z axis at z=d

Answer 1

Concepts and reason

The concept used to solve this problem is electric field.

Initially calculate the direction of electric field at any point on the z-axis. After that calculate the magnitude of the electric field along positive z axis. Finally, calculate the ball’s subsequent trajectory.

Fundamentals

The expression for electric field is,

$E = \frac{{kq}}{{{r^2}}}$

Here, $k = \frac{1}{{4\pi {\varepsilon _0}}}$ is an electrostatic constant, ${\varepsilon _0}$ is the vacuum permittivity, $q$ is the charge, and $r$ is the distance between the charge and point of observation of electric field.

The magnitude of force on a charge in electric field is,

$F = {q_0}E$

Here, ${q_0}$ is the charge on the particle in the electric field, and $E$ is the electric field.

The projection of any vector along the $z$ axis is,

$A\left( z \right) = A\cos \theta$

Here, $A$ is the magnitude of vector $\vec A$ , and $\theta$ is the angle the vector makes counterclockwise from the axis.

The Newton’s second law can be expressed as,

$\sum {\vec F = m\vec a}$

Here, $\sum {\vec F}$ is the vector sum of all the forces acting on the object, $m$ is the mass, and $\vec a$ is the acceleration of object.

The acceleration along $+ z$ axis is,

$\vec a = \frac{{{d^2}z}}{{d{t^2}}}$

Here, $z$ is the positions coordinate of the object, and $t$ is the time.

The equation of motion of a simple harmonic oscillation along $z$ axis is,

$\frac{{{d^2}z}}{{d{t^2}}} = - {\omega ^2}z$

Here, $\omega$ is the angular frequency.

In a right-angled triangle, the cosine function is,

$\cos \theta = \frac{{{\rm{adjacent}}}}{{{\rm{hypotenuse}}}}$

Pythagoras theorem gives the hypotenuse of right angled triangle as,

${\rm{hypotenuse}} = \sqrt {{{\left( {{\rm{adjacent}}} \right)}^2} + {{\left( {{\rm{opposite}}} \right)}^2}}$

(A)

Draw the diagram of the electric field $d\vec E$ due to a charge $dq$ on the uniformly charged ring with a radius $a$ and positive charge $q$ distributed evenly along its circumference. The magnitude of electric field $d\vec E$ is $dE$ . The position coordinate of any arbitrary point on $+ z$ axis is $z$ . The angle of electric field makes with $+ z$ axis in counterclockwise direction is $\theta$ . The tangent to a circle is perpendicular to the radius of the circle.

Substitute $z$ for ${\rm{adjacent}}$ , and $\sqrt {{z^2} + {a^2}}$ for ${\rm{hypotenuse}}$ in the cosine function equation $\cos \theta = \frac{{{\rm{adjacent}}}}{{{\rm{hypotenuse}}}}$ and solve for $\cos \theta$ .

$\cos \theta = \frac{z}{{\sqrt {{z^2} + {a^2}} }}$ …… (1)

Substitute $dE\left( z \right)$ for $A\left( z \right)$ , and $dE$ for $A$ in the equation $A\left( z \right) = A\cos \theta$ to get the electric field along $+ z$ axis.

$dE\left( z \right) = dE\cos \theta$ …… (2)

Here, $dE\left( z \right)$ is the electric field along $z$ axis.

Use the electric field equation.

$E = \frac{{kq}}{{{r^2}}}$

Substitute $\sqrt {{z^2} + {a^2}}$ for $r$ in the above equation and solve for the magnitude of the electric field due to the uniformly charged ring.

$\begin{array}{c}\\E = \frac{{kq}}{{{{\left( {\sqrt {{z^2} + {a^2}} } \right)}^2}}}\\\\ = \frac{{kq}}{{\left( {{z^2} + {a^2}} \right)}}\;\;{\rm{ }}.......{\rm{ (3)}}\\\end{array}$

Integrate the equation (2) to solve for the magnitude of electric field due to complete ring.

$\begin{array}{c}\\\int {d{E_z}} = \int {dE\cos \theta } \\\\E\left( z \right) = E\cos \theta \\\end{array}$

Here, $E\left( z \right)$ is the electric field along $z$ direction.

Substitute $\frac{z}{{\sqrt {{z^2} + {a^2}} }}$ for $\cos \theta$ from equation (1), and $\frac{{kq}}{{\left( {{z^2} + {a^2}} \right)}}$ for $E$ from the equation (3) in the above equation $E\left( z \right) = E\cos \theta$ .

$\begin{array}{c}\\E\left( z \right) = \left( {\frac{{kq}}{{\left( {{z^2} + {a^2}} \right)}}} \right)\left( {\frac{z}{{\left( {\sqrt {{z^2} + {a^2}} } \right)}}} \right)\\\\ = \frac{{kqz}}{{{{\left( {{z^2} + {a^2}} \right)}^{\frac{3}{2}}}}}\\\end{array}$

(B)

Use the electrostatic force equation.

$F = {q_0}E$

Here, ${q_0}$ is the charge on the ball, and $E$ is the magnitude of electric field along $z$ axis.

$\begin{array}{c}\\F = {q_0}\frac{{kqz}}{{{{\left( {{z^2} + {a^2}} \right)}^{\frac{3}{2}}}}}\\\\ = \frac{{kq{q_0}}}{{{{\left( {{z^2} + {a^2}} \right)}^{\frac{3}{2}}}}}z\\\end{array}$ Substitute $\frac{{kqz}}{{{{\left( {{z^2} + {a^2}} \right)}^{\frac{3}{2}}}}}$ for $E$ in the above equation $F = {q_0}E$ from the part (1).

…… (4)

Substitute $d$ for $z$ to find the force at any end point of the oscillations in the equation (4).

$F = \frac{{kq{q_0}}}{{{{\left( {{z^2} + {a^2}} \right)}^{\frac{3}{2}}}}}z$ .

$F = \frac{{kq{q_0}}}{{{{\left( {{d^2} + {a^2}} \right)}^{\frac{3}{2}}}}}d$

Use the approximation ${d^2} + {a^2} \approx {a^2}$ as $\left( {d < < a} \right)$ in the above equation.

$\begin{array}{c}\\F = \frac{{kq{q_0}}}{{{{\left( {{a^2}} \right)}^{\frac{3}{2}}}}}d\\\\ = \frac{{kq{q_0}}}{{{a^3}}}d\\\end{array}$

Therefore, force at any point along the $z$ axis can be approximated from the above result. Substitute $- z$ for $d$ in the above expression to solve for the force along $z$ axis.

$\vec F = - \frac{{kq{q_0}}}{{{a^3}}}z$

Use the Newton’s second law to solve for the acceleration of the ball.

$\sum {\vec F = m\vec a}$

Here, $\sum {\vec F}$ is the force acting on the ball, $m$ is the mass of the ball, and $\vec a$ is the acceleration of the ball.

Substitute $- \frac{{kq{q_0}}}{{{a^3}}}z$ for $\sum {\vec F}$ , and $\frac{{{d^2}z}}{{d{t^2}}}$ for $\vec a$ in the above equation and solve for the equation of motion.

$\begin{array}{c}\\ - \frac{{kq{q_0}}}{{{a^3}}}z = m\frac{{{d^2}z}}{{d{t^2}}}\\\\\frac{{{d^2}z}}{{d{t^2}}} = - \frac{{kq{q_0}}}{{m{a^3}}}z\\\end{array}$

Compare this equation of motion with equation of motion of the simple harmonic motion to solve for the angular frequency.

$\omega = \sqrt {\frac{{kq{q_0}}}{{m{a^3}}}}$

Ans:

The direction of electric field at any point on the z axis is parallel to the z axis.