Chapter 16, Problem 80QAP

To determine

(a)

The expression for electric field at point X.

Answer to Problem 80QAP

The expression for electric field at point X is, $E_x=9600.56\text{N/C}$

Explanation of Solution

Given:

Charge, $q_A=+20.0\text{nC=}20.0\times {\text{10}}^{-9}\text{C}$

Charge, $q_B=-8.00\text{nC=}-\text{8}{\text{.00×10}}^{\text{-9}}\text{C}$

Charge, $q_C=-10.0\text{nC=}-10.0\times {\text{10}}^{-9}\text{C}$

Distance, $s=10.0\text{cm=0}\text{.10m}$

Formula used:

The electric field is given by,
  $E=\frac{kq}{r^2}$

Where,
  $E$ =Electric field
  $q$ = Electric charges
  $r$ =Distance from charge

Calculation:

The electric field at point X is given by,
  $E_x=\frac{k{q}_A}{r_A^2}+\frac{k{q}_B}{r_B^2}+\frac{k{q}_C}{r_C^2}$

To determine

(b)

The expression for electric field at point Y.

Answer to Problem 80QAP

The expression for electric field at point Y is, $E_y=-11568.7\text{N/C}$

Explanation of Solution

Given:

Charge, $q_A=+20.0\text{nC=}20.0\times {\text{10}}^{-9}\text{C}$

Charge, $q_B=-8.00\text{nC=}-\text{8}{\text{.00×10}}^{\text{-9}}\text{C}$

Charge, $q_C=-10.0\text{nC=}-10.0\times {\text{10}}^{-9}\text{C}$

Distance, $s=10.0\text{cm=0}\text{.10m}$

Formula used:

The electric field is given by,
  $E=\frac{kq}{r^2}$

Where,
  $E$ =Electric field
  $q$ = Electric charges
  $r$ =Distance from charge

The electric field at point Y is given by,
  $E_y=\frac{k{q}_A}{r_{yA}^2}+\frac{k{q}_B}{r_{yB}^2}+\frac{k{q}_C}{r_{yC}^2}$

To determine

(c)

The expression for electric field at point Z.

Answer to Problem 80QAP

The expression for electric field at point Z is, $E_y=-11568.7\text{N/C}$

Explanation of Solution

Given:

Charge, $q_A=+20.0\text{nC=}20.0\times {\text{10}}^{-9}\text{C}$

Charge, $q_B=-8.00\text{nC=}-\text{8}{\text{.00×10}}^{\text{-9}}\text{C}$

Charge, $q_C=-10.0\text{nC=}-10.0\times {\text{10}}^{-9}\text{C}$

Distance, $s=10.0\text{cm=0}\text{.10m}$

Formula used:

The electric field is given by,
  $E=\frac{kq}{r^2}$

Where,
  $E$ =Electric field
  $q$ = Electric charges
  $r$ =Distance from charge

The electric field at point Z is given by,
  $E_z=\frac{k{q}_A}{r_{zA}^2}+\frac{k{q}_B}{r_{zB}^2}+\frac{k{q}_C}{r_{zC}^2}$

To determine

(d)

The electric field at point X, Y and Z.

Answer to Problem 80QAP

The electric field at point X, Y and Z are, $E_x=9600.56\text{N/C}$, $E_y=-11568.7\text{N/C}$ and $E_z=-9240.56\text{N/C}$

Explanation of Solution

Given:

Charge, $q_A=+20.0\text{nC=}20.0\times {\text{10}}^{-9}\text{C}$

Charge, $q_B=-8.00\text{nC=}-\text{8}{\text{.00×10}}^{\text{-9}}\text{C}$

Charge, $q_C=-10.0\text{nC=}-10.0\times {\text{10}}^{-9}\text{C}$

Distance, $s=10.0\text{cm=0}\text{.10m}$

Formula used:

The electric field is given by,
  $E=\frac{kq}{r^2}$

Where,
  $E$ =Electric field
  $q$ = Electric charges
  $r$ =Distance from charge

Calculation:

The electric field at point X is given by,
  $\begin{array}{l}{E}_x=\frac{k{q}_A}{r_A^2}+\frac{k{q}_B}{r_B^2}+\frac{k{q}_C}{r_C^2}\\ r_A=r_B=r_C=r=\frac{\sqrt{3}}{4}\times s=\frac{\sqrt{3}}{4}\times 0.1=0.04330\text{m}\\ E_x=\frac{k{q}_A}{r_A^2}+\frac{k{q}_B}{r_B^2}+\frac{k{q}_C}{r_C^2}=\frac{9\times {10}^9}{{\left(0.04330\right)}^2}(20.0\times {\text{10}}^{-9}-8.0\times {\text{10}}^{-9}-10.0\times {\text{10}}^{-9}\text{)}\\ E_x=9600.56\text{N/C}\end{array}$

The electric field at point Y is given by,
  $\begin{array}{l}{E}_y=\frac{k{q}_A}{r_{yA}^2}+\frac{k{q}_B}{r_{yB}^2}+\frac{k{q}_C}{r_{yC}^2}\\ r_{yA}=r_{yB}=\frac{s}{2}=5.0\text{cm=0}\text{.5m}\\ r_{yC}=\frac{\sqrt{3}}{2}\times s=\frac{\sqrt{3}}{4}\times 0.1=0.0866\text{m}\\ E_y=\frac{k{q}_A}{r_{yA}^2}+\frac{k{q}_B}{r_{yB}^2}+\frac{k{q}_C}{r_{yC}^2}\\ E_y=\frac{9\times {10}^9}{{\left(0.5\right)}^2}(20.0\times {\text{10}}^{-9}-8.0\times {\text{10}}^{-9})-\frac{9\times {10}^9\times 10.0\times {\text{10}}^{-9}}{{\left(0.0866\right)}^2}\\ E_y=-11568.7\text{N/C}\end{array}$

The electric field at point Z is given by,
  $\begin{array}{l}{E}_z=\frac{k{q}_A}{r_{zA}^2}+\frac{k{q}_B}{r_{zB}^2}+\frac{k{q}_C}{r_{zC}^2}\\ r_{zA}=r_{zC}=\frac{s}{2}=5.0\text{cm=0}\text{.5m}\\ r_{zB}=\frac{\sqrt{3}}{2}\times s=0.0866\text{m}\\ E_z=\frac{k{q}_A}{r_{zA}^2}+\frac{k{q}_B}{r_{zB}^2}+\frac{k{q}_C}{r_{zC}^2}\\ E_z=\frac{9\times {10}^9}{{\left(0.5\right)}^2}(20.0\times {\text{10}}^{-9}-10.0\times {\text{10}}^{-9})-\frac{9\times {10}^9\times 8.0\times {\text{10}}^{-9}}{{\left(0.0866\right)}^2}\\ E_z=-9240.56\text{N/C}\end{array}$