Chapter 24, Problem 67PQ

(a)

To determine

The magnitude of the electric field along the ring’s axis at a distance of $2.50\text{cm}$ from its centre.

Answer to Problem 67PQ

The magnitude of the electric field along the ring’s axis at a distance of $2.50\text{cm}$ from its centre is $5.86\times {10}^6\text{N/C}$.

Explanation of Solution

Write the equation for the electric field along the axis of a charged ring.

$E=\frac{kQy}{{\left(R^2+y^2\right)}^{3/2}}$

Here, $E$ is the electric field, $Q$ is the charge, $R$ is the radius and $y$ is the distance from the ring’s centre.

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $54.0\text{μC}$ for $Q$, $12.5\text{cm}$ for $R$ and $2.50\text{cm}$ for $y$ in the above equation to find $E$.

$\begin{array}{c}E=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(54.0\text{μC}\right)\left(2.50\text{cm}\right)}{{\left({\left(12.5\text{cm}\right)}^2+{\left(2.50\text{cm}\right)}^2\right)}^{3/2}}\\ =\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(\left(54.0\text{μC}\right)\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(\left(2.50\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}{{\left({\left(\left(12.5\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2+{\left(\left(2.50\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\right)}^{3/2}}\\ =5.86\times {10}^6\text{N/C}\end{array}$

Thus, the magnitude of the electric field along the ring’s axis at a distance of $2.50\text{cm}$ from its centre is $5.86\times {10}^6\text{N/C}$.

(b)

To determine

The magnitude of the electric field along the ring’s axis at a distance of $12.5\text{cm}$ from its centre.

Answer to Problem 67PQ

The magnitude of the electric field along the ring’s axis at a distance of $12.5\text{cm}$ from its centre is $1.10\times {10}^7\text{N/C}$.

Explanation of Solution

Write the equation for the electric field along the axis of a charged ring.

$E=\frac{kQy}{{\left(R^2+y^2\right)}^{3/2}}$

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $54.0\text{μC}$ for $Q$, $12.5\text{cm}$ for $R$ and $12.5\text{cm}$ for $y$ in the above equation to find $E$.

$\begin{array}{c}E=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(54.0\text{μC}\right)\left(12.5\text{cm}\right)}{{\left({\left(12.5\text{cm}\right)}^2+{\left(12.5\text{cm}\right)}^2\right)}^{3/2}}\\ =\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(\left(54.0\text{μC}\right)\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(\left(12.5\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}{{\left({\left(\left(12.5\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2+{\left(\left(12.5\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\right)}^{3/2}}\\ =1.10\times {10}^7\text{N/C}\end{array}$

Thus, the magnitude of the electric field along the ring’s axis at a distance of $12.5\text{cm}$ from its centre is $1.10\times {10}^7\text{N/C}$.

(c)

To determine

The magnitude of the electric field along the ring’s axis at a distance of $25.0\text{cm}$ from its centre.

Answer to Problem 67PQ

The magnitude of the electric field along the ring’s axis at a distance of $25.0\text{cm}$ from its centre is $5.56\times {10}^6\text{N/C}$.

Explanation of Solution

Write the equation for the electric field along the axis of a charged ring.

$E=\frac{kQy}{{\left(R^2+y^2\right)}^{3/2}}$

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $54.0\text{μC}$ for $Q$, $12.5\text{cm}$ for $R$ and $25.0\text{cm}$ for $y$ in the above equation to find $E$.

$\begin{array}{c}E=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(54.0\text{μC}\right)\left(25.0\text{cm}\right)}{{\left({\left(12.5\text{cm}\right)}^2+{\left(25.0\text{cm}\right)}^2\right)}^{3/2}}\\ =\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(\left(54.0\text{μC}\right)\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(\left(25.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}{{\left({\left(\left(12.5\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2+{\left(\left(25.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\right)}^{3/2}}\\ =5.56\times {10}^6\text{N/C}\end{array}$

Thus, the magnitude of the electric field along the ring’s axis at a distance of $25.0\text{cm}$ from its centre is $5.56\times {10}^6\text{N/C}$.

(d)

To determine

The magnitude of the electric field along the ring’s axis at a distance of $2.00\text{m}$ from its centre.

Answer to Problem 67PQ

The magnitude of the electric field along the ring’s axis at a distance of $2.00\text{m}$ from its centre is $1.21\times {10}^5\text{N/C}$.

Explanation of Solution

Write the equation for the electric field along the axis of a charged ring.

$E=\frac{kQy}{{\left(R^2+y^2\right)}^{3/2}}$

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $54.0\text{μC}$ for $Q$, $12.5\text{cm}$ for $R$ and $2.00\text{m}$ for $y$ in the above equation to find $E$.

$\begin{array}{c}E=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(54.0\text{μC}\right)\left(2.00\text{m}\right)}{{\left({\left(12.5\text{cm}\right)}^2+{\left(2.00\text{m}\right)}^2\right)}^{3/2}}\\ =\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(\left(54.0\text{μC}\right)\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)\left(2.00\text{m}\right)}{{\left({\left(\left(12.5\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2+{\left(2.00\text{m}\right)}^2\right)}^{3/2}}\\ =1.21\times {10}^5\text{N/C}\end{array}$

Thus, the magnitude of the electric field along the ring’s axis at a distance of $2.00\text{m}$ from its centre is $1.21\times {10}^5\text{N/C}$.