Chapter 20, Problem 65A

a.

To determine

The magnitude of forces on the charge $q_T$ .

Answer to Problem 65A

The magnitude of forces on the charge $q_T$ area

  $F_A=370\text{ N}$ towards $q_T$

  $F_B=92\text{ N}$ away from $q_T$ .

Explanation of Solution

Given:

Charge, $q_T=7.\text{2 μC}$

Charge, $q_A=3.6\mu C$

Charge, $q_B=-6.6\mu C$

Distance between A and T, $d_1=0.025\text{ m}$

Distance between B and T, $d_2=0.068\text{ m}$

Formula Used:

  $F=\frac{k{Q}_1{Q}_2}{d^2}$

F is the force

kis the constant

Q represents the charges

d is the distance between the charges.

Calculations:

  $F=\frac{k{Q}_1{Q}_2}{d^2}$

For A and T, substitute the values

  $F_A=\frac{\left(9.9\times {10}^9\text{ N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(7.2\text{ μC}\right)\left(3.6\text{ μC}\right)}{{\left(0.025\text{ m}\right)}^2}$

  $F_A=370\text{ N}$

For B and T substitute the values

  $F_B=\frac{\left(9.9\times {10}^9\text{ N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(\text{7}\text{.2 μC}\right)\left(\text{-6}\text{.6 μC}\right)}{{\left(0.068\text{ m}\right)}^2}$

  $F_B=92\text{ N}$

Conclusion:

The magnitude of forces on the charge $q_T$ are-

  $F_A=370\text{ N}$ towards $q_T$

  $F_B=92\text{ N}$ away from $q_T$ .

b.

To determine

To Sketch: A force diagram.

Explanation of Solution

Introduction:

When charged particles are placed in the vicinity of each other they experience a force known as electrostatic force. Same charge particles experience repulsive force while oppositely charged particles experience an attractive force.

The forces on the charge are shown below

  $F_A=370\text{ N}$ towards $q_T$

  $F_B=92\text{ N}$ away from $q_T$ .

Conclusion:

  $F_A=370\text{ N}$ towards $q_T$

  $F_B=92\text{ N}$ away from $q_T$ .

c.

To determine

The resultant force graphically.

Explanation of Solution

Introduction:

When charged particles are placed in the vicinity of each other they experience a force known as electrostatic force. Same charge particles experience repulsive force while oppositely charged particles experience an attractive force.

The forces on the charge are shown below

  $F_A=370\text{ N}$ towards $q_T$

  $F_B=92\text{ N}$ away from $q_T$ .

Vector addition gives the resultant force.

  $\begin{array}{c}{F}_R=\sqrt{F_A{}^2+F_B{}^2}\\ =\sqrt{370{\text{ N}}^2+92{\text{ N}}^2}\\ =380\text{N}\end{array}$

Conclusion:

  $F_A=370\text{ N}$ toward $q_T$

  $F_B=92\text{ N}$ away from $q_T$

  $F_R=380\text{N}$