Chapter 18, Problem 41TP

To determine

(a)

(i)

The characteristics of field diagram, which indicates the objects R and T sign of chargesare positive and for object S is negative.

(ii)

The characteristics of field diagram, which indicates the objects R and T magnitude are equal and for object S is twice than that of object R and T.

Answer to Problem 41TP

1. The characteristics of field diagram indicates the sign of the objects R and T which are positive and for object S as negative, which is described by the direction of field vectors.
2. The characteristics of field diagramindicates the magnitude of the objects R and T as equal and for object S as twice than that of object R and T.The magnitude is described by the distance of field vectors.

Explanation of Solution

Introduction:

The lines of forces are imaginary which signify amplitude and direction of an electric field at any point. The lines of forces direction could be in opposite for positive and negative charged particles.

1. The characteristics of field diagram indicates the magnitude of the objects R and T as equal and for object S as twice than that of object R and T. The magnitude is described by the distance of field vectors.
The electric field in between the object S and T is the total electric field of object R,S and T.
2. The field magnitude is given by

  $E=\left(k\frac{q}{r^2}\right)$

The R and T field magnitude are equal.

The vector length is same for the objects closest to R and T. And it starts from R and T in the same distance of about 4 units. Hence, the R and T charges could be same.

But the vector length of object S is same as of object R and T and it starts from larger distance like 6 units when compared to R and T.

The field vector starts from 4 units and field vector starts from 6 units are same and it is given as

  $\begin{array}{l}{E}_{R(4units)}=E_{S(6units)}\\ k\frac{q_R}{{\left(d_R\right)}^2}=k\frac{q_S}{{\left(d_S\right)}^2}\end{array}$

Where $q_R$ and $q_S$ are charges, $d_R$ and $d_S$ distance from object R and S.

Substitute the corresponding units for distances

  $\begin{array}{l}k\frac{q_R}{{\left(4\right)}^2}=k\frac{q_S}{{\left(6\right)}^2}\\ \frac{q_S}{q_R}=\frac{6^2}{4^2}\\ =\frac{36}{16}\\ =2.25\end{array}$

Hence the charge $q_S$ is twice of charge $q_R$

Conclusion:

Therefore, by the characteristics of field diagram, which indicates the objects R and T sign of charges are positive and for object S is negative which is described by the direction of field vectors.

The characteristics of field diagram, which indicates the objects R and T magnitude, are equal and for object S is twice than that of object R and T which is described by the distance of field vectors.

To determine

(b)

To Plot:

The graph of electric field E as a function of position of xalong the x-axis.

Answer to Problem 41TP

The graph of electric field E in position of x along the x- axis is plotted.

Explanation of Solution

Introduction:

The lines of forces are imaginary lines which signify amplitude and direction of an electric field at any point. The lines of forces direction could be opposite for positive and negative charged particles.

The graph of electric field E in position of x along the x-axis is plotted below.

  

To determine

(c)

An expression for electric field E in terms of charge, distance and fundamental constants.

Answer to Problem 41TP

The expression for electric field E in terms of charge, distance and fundamental constants is written as $E=kq\left(\frac{1}{{\left(d-x\right)}^2}+\frac{2}{{\left(d-x\right)}^2}+\frac{-1}{{\left(d+x\right)}^2}\right)$.

Explanation of Solution

Introduction:

The lines of forces are imaginary lines which signify amplitude and direction of an electric field at any point. The lines of forces direction could be in opposite for positive and negative charged particles.The electric field in between the object S and T is the total electric field of object R, S and T.

The lines of forces direction could be in opposite for positive and negative charged particles. The lines of forces are imaginary lines which signify amplitude and direction of an electric field at any point. The electric field of object R at distance x from object S is given as,

  $E=k\frac{-q}{{\left(d+x\right)}^{2^{}}}$

The electric field of object S at distance x from object S is given as,

  $E=k\frac{2q}{{\left(x\right)}^{2^{}}}$

The electric field of object T at distance x from object S is given as,

  $E=k\frac{q}{{\left(d-x\right)}^{2^{}}}$

The total electric field is given as

  $E=kq\left(\frac{1}{{\left(d-x\right)}^2}+\frac{2}{{\left(d-x\right)}^2}+\frac{-1}{{\left(d+x\right)}^2}\right)$

Hence the net electric field is $E=kq\left(\frac{1}{{\left(d-x\right)}^2}+\frac{2}{{\left(d-x\right)}^2}+\frac{-1}{{\left(d+x\right)}^2}\right)$

Conclusion:

Therefore,the expression for electric field E in terms of charge, distance and fundamental constants is written as $E=kq\left(\frac{1}{{\left(d-x\right)}^2}+\frac{2}{{\left(d-x\right)}^2}+\frac{-1}{{\left(d+x\right)}^2}\right)$.

To determine

(d)

Whether the given statement is true or false by using the above representations.

Answer to Problem 41TP

The given statement is false.

Explanation of Solution

Introduction:

The lines of forces are imaginary lines which signify amplitude and direction of an electric field at any point. The electric field in between the object S and T is the total electric field of object R,S and T.

The expression for electric field E in terms of charge, distance and fundamental constants is written as $E=kq\left(\frac{1}{{\left(d-x\right)}^2}+\frac{2}{{\left(d-x\right)}^2}+\frac{-1}{{\left(d+x\right)}^2}\right)$.

The statement is incorrect since the vector length in the region is non-zero.

From the equation, it is stated that the object R denominator is always greater than that the denominator of object T. But the numerators are same. Hence, the magnitude of S is smaller than that of T. Also, if the second term is positive, the whole term is positive.

Conclusion:

Therefore,the statement is not correct.