Chapter 15, Problem 18P

(a) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in Figure P15.10.

(b) If a charge of −2.00 μC is placed at this point, what are the magnitude and direction of the force on it?

To determine

The electric field at a point 1.00 cm from $q_2$.

Answer to Problem 18P

Solution: The electric field at a point 1.00 cm from $q_2$ is $2.00\times {10}^7\text{N/C}$.

Explanation of Solution

The force diagram is given by,

In the above diagram,

• $F_{q_1}$ is the force due to $q_1$.
• $F_{q_2}$ is the force due to $q_2$.
• $F_{q_3}$ is the force due to $q_3$.
• $E_1$, $E_2$ and $E_3$ are the electric field corresponding to the three charges.

Formula to calculate the electric is given by,

$E=\frac{k_{e}Q}{r^2}$

• $k_e$ is the Coulomb constant.
• Q is the charge.
• r is the distance of separation.

Formula to calculate the electric field at a point 1.00 cm from $q_2$ is,

$\begin{array}{c}E=E_1+E_3-E_2\\ =\left(\frac{k_e{q}_1}{r_1^2}\right)+\left(\frac{k_e{q}_3}{r_3^2}\right)-\left(\frac{k_e{q}_2}{r_2^2}\right)\end{array}$

• $r_1$ is the distance between $q_1$ and the point.
• $r_2$ is the distance between $q_2$ and the point.
• $r_3$ is the distance between $q_3$ and the point.

Substitute $8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k_e$, $6\text{μC}$ for $q_1$, $1.5\text{μC}$ for $q_2$, $2\text{μC}$ for $q_3$, 2.00 cm for $r_1$, 3.00 cm for $r_3$ and 1.00 cm for $r_2$.

$\begin{array}{c}E=\left(8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left[\left(\frac{6\text{μC}}{{\left(2.00\text{cm}\right)}^2}\right)+\left(\frac{2\text{μC}}{{\left(3.00\text{cm}\right)}^2}\right)-\left(\frac{1.5\text{μC}}{{\left(1.00\text{cm}\right)}^2}\right)\right]\\ =\left(8.99\times {10}^9\text{N}{\text{.m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left[\left(\frac{6\times {10}^{-6}\text{C}}{{\left(1.00\times {10}^{-2}\text{m}\right)}^2}\right)+\left(\frac{2\times {10}^{-6}\text{C}}{{\left(3.00\times {10}^{-2}\text{m}\right)}^2}\right)-\left(\frac{1.5\times {10}^{-6}\text{C}}{{\left(1.00\times {10}^{-2}\text{m}\right)}^2}\right)\right]\\ =2.00\times {10}^7\text{N/C}\end{array}$

Conclusion:

The electric field at a point 1.00 cm from $q_2$ is $2.00\times {10}^7\text{N/C}$.

To determine

The force on the charge placed at a point 1.00 cm from $q_2$.

Answer to Problem 18P

Solution: The force on the charge placed at a point 1.00 cm from $q_2$ is towards the left and has magnitude of 40.0 N.

Explanation of Solution

Given info: $-2.00\text{μC}$ charge is placed at a point 1.00 cm from $q_2$.

Formula to calculate the force on the charge placed at a point 1.00 cm from $q_2$. is,

$F=\left|q\right|E$

• q is the charge placed at the point.

Substitute $2.00\times {10}^7\text{N/C}$ for E and $-2\text{μC}$ for q.

$\begin{array}{c}F=\left|\left(-2\text{μC}\right)\right|\left(2.00\times {10}^7\text{N/C}\right)\\ =\left(2\times {10}^{-6}\text{C}\right)\left(2.00\times {10}^7\text{N/C}\right)\\ =40.0\text{N}\end{array}$

Conclusion:

The force on the charge placed at a point 1.00 cm from $q_2$ is towards the left and has magnitude of 40.0 N