Chapter 26, Problem 64PQ

(a)

To determine

The finite values of y for which the electric field is zero.

Answer to Problem 64PQ

The finite values of y for which the electric field is zero is $-2.05\text{m}$.

Explanation of Solution

Write an expression for the total electric field equal to zero.

  $E_y=\left(k\frac{q_A}{y^2}+k\frac{q_B}{{\left(y-y\text{'}\right)}^2}\right)$                                                                                      (I)

Here, $E_y$ is the electric field, $k$ is the culoumb constant, $q_A$ is the first charge, $q_B$ is the second charge, $y$ is the position where total electric field is zero and $y\text{'}$ is the position of the second charge.

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $1.00\text{nC}$ for $q_A$, $-3.00\text{nC}$ for $q_B$, $1.50\text{m}$ for $y\text{'}$ and $0$ for $E_y$ in equation (I).

  $\begin{array}{c}0=\left(\begin{array}{l}\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(\left(1.00\text{nC}\right)\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{y^2}+\\ \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(\left(-3.00\text{nC}\right)\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{{\left(y-1.50\text{m}\right)}^2}\end{array}\right)\\ =\left(\begin{array}{l}\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(1.00\times {10}^{-9}\text{C}\right)}{y^2}+\\ \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(-1.00\times {10}^{-9}\text{C}\right)}{{\left(y-1.50\text{m}\right)}^2}\end{array}\right)\\ =\left(\frac{\left(1.00\right)}{y^2}-\frac{\left(3.00\right)}{{\left(y-1.50\right)}^2}\right)\end{array}$                                          (II)

Rearrange the equation (II).

  $\begin{array}{c}0=\frac{\left(1.00\right){\left(y-1.50\right)}^2-\left(3.00\right)y^2}{y^2{\left(y-1.50\right)}^2}\\ =\left(1.00\right){\left(y-1.50\right)}^2-\left(3.00\right)y^2\\ =2.00y^2+3.00y-2.25\end{array}$                                                                       (III)

Calculate the value of $y$ from equation (III).

    $\begin{array}{c}y=\frac{-3.00\pm \sqrt{{\left(3.00\right)}^2-4\left(2.00\right)\left(-2.25\right)}}{2\left(2.00\right)}\\ =\frac{-3.00\pm \sqrt{27.0}}{2\left(2.00\right)}\\ =0.55\text{m}\text{or}-2.05\text{m}\end{array}$

Thus, the finite value of y for which the electric field is zero is $-2.05\text{m}$ since the positive root is not physically valid.

(b)

To determine

The finite values of y for which the electric potential is zero.

Answer to Problem 64PQ

The finite values of y for which the electric potential is zero are $0.375\text{m}$ and $-0.750\text{m}$.

Explanation of Solution

Write an expression for the total electric field equal to zero.

  $V=\left(k\frac{q_A}{y}+k\frac{q_B}{\left(y\text{'}-y\right)}\right)$                                                                                       (IV)

Here, $V$ is the electric potential.

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $1.00\text{nC}$ for $q_A$, $-3.00\text{nC}$ for $q_B$, $1.50\text{m}$ for $y\text{'}$ and $0$ for $V$ in equation (IV).

  $\begin{array}{c}0=\left(\begin{array}{l}\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(\left(1.00\text{nC}\right)\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{y}+\\ \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(\left(-3.00\text{nC}\right)\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{\left(1.50\text{m}-y\right)}\end{array}\right)\\ =\left(\begin{array}{l}\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(1.00\times {10}^{-9}\text{C}\right)}{y}+\\ \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(-3.00\times {10}^{-9}\text{C}\right)}{\left(1.50\text{m}-y\right)}\end{array}\right)\\ =-4y+1.50\text{m}\end{array}$                                          (V)

Rearraneg the equation (II).

  $4y=1.50\text{m}$                                                                                                         (VI)

Calculate the value of $y$ from equation (VI).

    $\begin{array}{c}y=\frac{1.50\text{m}}{4}\\ =0.375\text{m}\end{array}$

For y<0; Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $1.00\text{nC}$ for $q_A$, $-3.00\text{nC}$ for $q_B$, $1.50\text{m}$ for $y\text{'}$ and $0$ for $V$ in equation (IV).

  $\begin{array}{c}0=\left(\begin{array}{l}\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(\left(1.00\text{nC}\right)\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{-y}+\\ \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(\left(-3.00\text{nC}\right)\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{\left(1.50\text{m}-y\right)}\end{array}\right)\\ =\left(\begin{array}{l}\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(1.00\times {10}^{-9}\text{C}\right)}{-y}+\\ \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(-3.00\times {10}^{-9}\text{C}\right)}{\left(1.50\text{m}-y\right)}\end{array}\right)\\ =2y+1.50\text{m}\end{array}$                                        (VII)

Rearraneg the equation (VII).

  $2y=-1.50\text{m}$                                                                                                   (VIII)

Calculate the value of $y$ from equation (VIII).

    $\begin{array}{c}y=\frac{-1.50\text{m}}{2}\\ =0.750\text{m}\end{array}$

Thus, the finite values of y for which the electric potential is zero are $0.375\text{m}$ and $-0.750\text{m}$.