Q6Q

Question

In Fig. 22-27, two identical circular non-conducting rings are centered on the same line with their planes perpendicular to the line. Each ring has charge that is uniformly distributed along its circumference. The rings each produce electric fields at points along the line. For three situations, the charges on rings A and B are, respectively, (1) $q_{0}$ and $q_{0}$ , (2) $- q_{0}$ and $- q_{0}$ , and (3) $- q_{0}$ and. $q_{0}$ Rank the situations according to the magnitude of the net electric field at (a) point $P_{1}$ midway between the rings, (b) point $P_{2}$ at the center of ring B, and (c) point $P_{3}$ to the right of ring B, greatest first.

Step-by-Step Solution

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Answer

a)The rank of the situations according to the magnitude of net field at point $P_{1}$ is $| E_{(3)} | > | E_{(1)} | = | E_{(2)} |$ .

b) The rank of the situations according to the magnitude of net field at point $P_{2}$ is $| E_{(1)} | = | E_{(2)} | = | E_{(3)} |$ .

c) The rank of the situations according to the magnitude of net field at point $P_{3}$ is $| E_{(1)} | = | E_{(2)} | > | E_{(3)} |$ .

1Step 1: Understanding the concept of electric field of a ring

The electric field is directed towards the negative charge and away from the positive charge. If the right direction is said to be positive, the net field can be calculated at all the given axial points using the field formula.

The electric field of a ring at an axial point, $E = \frac{k q x}{{(x^{2} + a^{2})}^{3 / 2}}$ (i)

where, $x$ is the distance of the point from the center of the ring that lies on the axis and is the radius of the ring.

2Step 2: a) Calculation of the rank according to magnitude of electric fields at point P 1

Let, be the distance between the two rings be $d$ .

As the point is located in between the two rings $B$ , then the net electric filed due to the charges $q_{0} and q_{0}$ at the midway between the rings can be given using equation (i) as follows:

(Since, both the charges are positive, the field is radially away from them)

$\begin{matrix} E_{(1)} = \frac{8 k q (d / 2)}{{(d^{2} + 4 a^{2})}^{3 / 2}} - \frac{8 k q (d / 2)}{{(d^{2} + 4 a^{2})}^{3 / 2}} (As the point is at distance d/2 from each ring.) \\ | E_{(1)} | = 0 \end{matrix}$

Now, the net electric filed due to the charges at $- q_{0} and - q_{0}$ the midway between the rings can be given using equation (i) as follows:

$\begin{matrix} E_{(2)} = - \frac{8 k q (d / 2)}{{(d^{2} + 4 a^{2})}^{3 / 2}} + \frac{8 k q (d / 2)}{{(d^{2} + 4 a^{2})}^{3 / 2}} (As the point is at distance d/2 from each ring.) \\ | E_{(2)} | = 0 \end{matrix}$

Now, the net electric filed due to the charges $- q_{0} and q_{0}$ at the midway between the rings can be given using equation (i) as follows:

$\begin{matrix} E_{(3)} = - \frac{8 k q (d / 2)}{{(d^{2} + 4 a^{2})}^{3 / 2}} - \frac{8 k q (d / 2)}{{(d^{2} + 4 a^{2})}^{3 / 2}} (As the point is at distance d/2 from each ring.) \\ = - \frac{8 k q d}{{(d^{2} + 4 a^{2})}^{3 / 2}} \\ | E_{(3)} | = \frac{8 k q d}{{(d^{2} + 4 a^{2})}^{3 / 2}} \end{matrix}$

Hence, the required rank is. $| E_{(3)} | > | E_{(1)} | = | E_{(2)} |$

3Step 3: b) Calculation of the rank according to magnitude of electric fields at point P 2

Let, be the distance between the two rings be $d$ .

As the point is located at the center of ring B, then the net electric filed due to the charges $q_{0} and q_{0}$ at the center of ring B can be given using equation (i) as follows:

(Since, both the charges are positive, the field is radially away from them)

$\begin{matrix} E_{(1)} = \frac{k q d}{{(d^{2} + a^{2})}^{3 / 2}} - \frac{k q (0)}{{(0^{2} + a^{2})}^{3 / 2}} \\ | E_{(1)} | = \frac{k q d}{{(d^{2} + a^{2})}^{3 / 2}} \end{matrix}$

Now, the net electric filed due to the charges $- q_{0} and - q_{0}$ at thecenter of ring B can be given using equation (i) as follows: (Since, both the charges are positive, the field is radially towards them)

$\begin{matrix} E_{(2)} = - \frac{k q d}{{(d^{2} + a^{2})}^{3 / 2}} - \frac{k q (0)}{{(0^{2} + a^{2})}^{3 / 2}} \\ | E_{(2)} | = \frac{k q d}{{(d^{2} + a^{2})}^{3 / 2}} \end{matrix}$

Now, the net electric filed due to the charges $- q_{0} and q_{0}$ at the center of ring B can be given using equation (i) as follows:

$\begin{matrix} E_{(3)} = - \frac{k q d}{{(d^{2} + a^{2})}^{3 / 2}} + \frac{k q (0)}{{(0^{2} + a^{2})}^{3 / 2}} \\ | E_{(3)} | = \frac{k q d}{{(d^{2} + a^{2})}^{3 / 2}} \end{matrix}$

Hence, the required rank is $| E_{(1)} | = | E_{(2)} | = | E_{(3)} |$ .

4Step 4: c) Calculation of the rank according to magnitude of electric fields at point P 3

Let, be the distance between the two rings beand the point be at a distance $x$ from ring 2. Thus, the point distance from ring A will be $(d + x)$ .

As the point is located right of both rings, then the net electric filed due to the charges $q_{0} and q_{0}$ at the right of both rings can be given using equation (i) as follows:

(Since, both the charges are positive, the field is radially away from them)

$| E_{(1)} | = \frac{k q (d + x)}{{({(d + x)}^{2} + a^{2})}^{3 / 2}} + \frac{k q x}{{(x^{2} + a^{2})}^{3 / 2}}$

Now, the net electric filed due to the charges $- q_{0} and - q_{0}$ at the right of both rings can be given using equation (i) as follows:

(Since, both the charges are positive, the field is radially towards them)

$\begin{matrix} E_{(2)} = - \frac{k q (d + x)}{{({(d + x)}^{2} + a^{2})}^{3 / 2}} - \frac{k q x}{{(x^{2} + a^{2})}^{3 / 2}} \\ | E_{(2)} | = \frac{k q (d + x)}{{({(d + x)}^{2} + a^{2})}^{3 / 2}} + \frac{k q x}{{(x^{2} + a^{2})}^{3 / 2}} \end{matrix}$

Now, the net electric filed due to the charges $- q_{0} and q_{0}$ at the right of both rings can be given using equation (i) as follows:

$\begin{matrix} E_{(3)} = - \frac{k q (d + x)}{{({(d + x)}^{2} + a^{2})}^{3 / 2}} + \frac{k q (x)}{{(x^{2} + a^{2})}^{3 / 2}} \\ | E_{(3)} | = \frac{k q (d + x)}{{({(d + x)}^{2} + a^{2})}^{3 / 2}} - \frac{k q (x)}{{(x^{2} + a^{2})}^{3 / 2}} \end{matrix}$

Hence, the required rank is $| E_{(1)} | = | E_{(2)} | > | E_{(3)} |$ .

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