Chapter 18, Problem 19PE

To determine

(a)

To find:

The direction of the electric field at the center of the given square.

Answer to Problem 19PE

Electric field at the center of the square is in upward direction.

Explanation of Solution

Given:

Four charges q_a, q_b, q_c, q_d placed at each corner of the square with side 5 cmhaving magnitude q_a= q_b = -1.00 μC, q_c= q_d = +1.00 μC.

Electric field is something that is produced by the charge around it and another charge placed in it experiences a force. Charges of same sign repel each other and the charges of opposite sign attract each other.

In the given fig., place a test charge q at the center of the square. Charges q_a, q_battract the test charge with same because both the charges are negative in nature and having same magnitude. And the charges q_c, q_drepel the test charge with same force as the charges are of positive in nature and have the same magnitude.

Conclusion:

The direction of the resultant electric field at the center of the square is in upward direction due to entire system.

To determine

(b)

To calculate:

The magnitude of the electric field at the location of q.

Answer to Problem 19PE

  $E_R=13.58\times {10}^6\text{N/C}$

Explanation of Solution

Given:

Side of the square = 5 cm

  $\left|q_a\right|=\left|q_b\right| =\left|q_c\right|=\left|q_d\right|=1.00\mu C.$

Formula used:

  $E=\frac{1}{4\pi {\epsilon }_o}\frac{Q}{r^2}$

ε_ο = permittivity of free space,

E = intensity of the electric field,

q = given charge,

r = distance between the charge q and the point where electric field needs to be calculated.

Calculation:

  $\begin{array}{c}\text{Diagonal of the square = }\sqrt{2}s\\ =\sqrt{2}\times 5\times {10}^{-2}=0.07\text{m}\\ d=\frac{0.07}{2}=0.035\text{m}\end{array}$

The electric field at q due to the charge q_a and q_b

  $E_a=E_d=E_1(\text{let})$

  $\begin{array}{l}{E}_1=\frac{9\times {10}^9\times 1\times {10}^{-6}}{{\left(0.035\right)}^2}\\ =7.35\times {10}^6\text{N/C}\end{array}$

  $\begin{array}{l}{E}_b=E_c=E_2(\text{let})\\ E_2=\frac{9\times {10}^9\times 1\times {10}^{-6}}{{\left(0.035\right)}^2}\\ =7.35\times {10}^6\text{N/C}\end{array}$

  $\begin{array}{l}{E}_R=\sqrt{E_{{}_1}^2+E_2^2+2E_1{E}_2\cos \theta }\\ \theta ={45}^{{\rm O}}\\ \end{array}$

On putting the values,

  $\begin{array}{l}{E}_R=\sqrt{{\left(7.35\times {10}^6\right)}^2+{\left(7.35\times {10}^6\right)}^2+2\times 7.35\times {10}^6\times 7.35\times {10}^6\cos {45}^{{\rm O}}}\\ E_R=\sqrt{{\left(7.35\times {10}^6\right)}^2+{\left(7.35\times {10}^6\right)}^2+2\times 7.35\times {10}^6\times 7.35\times {10}^6\cos {45}^{{\rm O}}}\\ E_R=\sqrt{{\left(7.35\times {10}^6\right)}^2\left(1+1+2\cos {45}^{{\rm O}}\right)}\\ E_R=7.35\times {10}^6\sqrt{\left(1+1+2\cos {45}^{{\rm O}}\right)}\\ E_R=7.35\times {10}^6\times 1.85\\ E_R=13.58\times {10}^6\text{N/C}\end{array}$

Conclusion:

The magnitude of the electric field at the location of q is $E_R=13.58\times {10}^6\text{N/C}$