Chapter 24, Problem 15PQ

To determine

The position of the third charge such that the electric field at the origin is zero.

Answer to Problem 15PQ

The position of the third charge such that the electric field at the origin is zero is $-0.0986\text{m}$.

Explanation of Solution

Write the equation for the electric field chargeq.

    $E_q=k\frac{q}{r^2}\widehat{i}$                                                                                   (I)

Here, $E_q$ is the electric field, $q$ is the charge, $k$ is the Coulomb constant and $r$ is the distance of the charge from the origin.

Write the equation for the electric field due to charge 2q.

    $E_{2q}=k\frac{2q}{r^2}\widehat{i}$                                                                                (II)

Here, $E_{2q}$ is the electric field due to second particle.

Write the equation for the electric field due to charge 3q.

    $E_{3q}=k\frac{3q}{x^2}\widehat{i}$                                                                                (III)

Here, $E_{2q}$ is the electric field due to second particle and $x$ is the position of the third charge such that the electric field at the origin is zero.

Write the equation for the net electric field.

    $E_{net}=E_q+E_{2q}$                                                                             (IV)

Write the condition that the electric field at the origin is zero using equation (III).

    $\begin{array}{c}{E}_{\text{final net}}=0\\ E_{net}+E_{3q}=0\\ E_{net}+k\frac{3q}{x^2}\widehat{i}=0\end{array}$                                                                             (V)

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $2.57\text{μC}$ for $q$ and $0.0569\text{m}$ for $r$ in equation (I) to find $E_q$.

    $\begin{array}{c}{E}_q=\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(2.57\text{μC}\right)}{{\left(0.0569\text{m}\right)}^2}\widehat{i}\\ =\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(2.57\text{μC}\right)\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{{\left(0.0569\text{m}\right)}^2}\widehat{i}\\ =7.136\times {10}^6\widehat{i}\text{N/C}\end{array}$

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $5.14\text{μC}$ for $2q$ and $0.0569\text{m}$ for $r$ in equation (II) to find $E_{2q}$.

    $\begin{array}{c}{E}_{2q}=\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(5.14\text{μC}\right)}{{\left(0.0569\text{m}\right)}^2}\left(-\widehat{i}\right)\\ =\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(5.14\text{μC}\right)\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{{\left(0.0569\text{m}\right)}^2}\left(-\widehat{i}\right)\\ =-14.272\times {10}^6\widehat{i}\text{N/C}\end{array}$

Substitute $7.136\times {10}^6\widehat{i}\text{N/C}$ for $E_q$ and $-14.272\times {10}^6\widehat{i}\text{N/C}$ for $E_{2q}$ in equation (IV) to find $E_{net}$.

    $\begin{array}{c}{E}_{net}=7.136\times {10}^6\widehat{i}\text{N/C+}\left(-14.272\times {10}^6\widehat{i}\text{N/C}\right)\\ =-7.136\times {10}^6\widehat{i}\text{N/C}\end{array}$

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $7.71\text{μC}$ for $3q$, $-7.136\times {10}^6\widehat{i}\text{N/C}$ for $E_{net}$ in equation (V) to solve for x.

    $\begin{array}{c}\left(-7.136\times {10}^6\widehat{i}\text{N/C}\right)+\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(7.71\text{μC}\right)}{x^2}\widehat{i}=0\\ x=\pm 0.0986\end{array}$

The direction of electric field must be in the positive x direction. Thus, the particle must be in the negative x axis.

Thus, the position of the third charge such that the electric field at the origin is zero is $-0.0986\text{m}$.