91影视

1. The magnitude of the electric field of an electric dipole at points along the dipole axis based on approximations is $E_{appr}$.
2. The actual magnitude of the electric field is$E_{act}$.
3. The distance of the point P from the dipole center, $z=5d$where, d is the separation between the two charge particles of a dipole.

An electric dipole is defined as a system of two opposite charges separated by a certain distance. This dipole thus creates a net electric field at any given point in the system due to the two electric fields produced by the two charges. Here, the point is along the axis on the two charges of the given dipole system lies. Again, the direction of the field is determined by the direction of the force. So, the field due to the two charges being in opposite directions will produce a net field that is given by the difference between the magnitudes of their fields. Now, relating it to the derived electric field based on certain approximations, the required ratio is determined.

Formula:

The magnitude of an electric field at a point due to a point charge,

$E=\frac{1}{2蟺{蔚}_o}\frac{q}{r^2}$ (i)

where,$k=\frac{1}{4蟺{蔚}_0}=9脳{10}^9{\text{N.m}}^2/{\text{C}}^2$is the constant value, R = The distance of field point from the charge, q= charge of the particle.

According to equation 22-8, the electric field due to an electric dipole with approximation $(d/2z鈮�1)$is given as follows:

$E=\frac{1}{2蟺{蔚}_o}\frac{qd}{z^3}$

鈥︹�︹�︹�︹�︹�︹�︹�︹�︹��.. (I)

Again, according to equation 22-9, the electric field due to an electric dipole with approximation$(d/2z鈮�1)$using dipole moment$(p=qd)$is given as follows:

$E=\frac{1}{2蟺{蔚}_o}\frac{p}{z^3}$

Now, as per the given data, the distance of the point P from negative charge of the dipole will be given as:

$\begin{array}{c}{r}_1=5d鈭�d/2\\ =4.5d\end{array}$

And, the distance of the point P from the positive charge will be given as:

$\begin{array}{c}{r}_2=5d+d/2\\ =4.5d\end{array}$

Now, the field direction due to two equal and opposite charges at that point P will be in opposite direction, thus the value of the magnitude of the actual electric field $E_{act}$is given using equation (i) as:

$\begin{array}{c}{E}_{act}=\frac{1}{4蟺{蔚}_o}\frac{q}{{r_1}^2}鈭�\frac{1}{4蟺{蔚}_o}\frac{q}{{r_2}^2}\\ =\frac{1}{4蟺{蔚}_o}\frac{q}{4.5d^2}鈭�\frac{1}{4蟺{蔚}_o}\frac{q}{5.5d^2}\\ =\frac{160}{9801}\frac{q}{4蟺{蔚}_o{d}^2}.\end{array}$ 鈥︹�︹�︹�︹�︹�︹�︹�︹�︹��.. (II)

And now, using the given data, the magnitude of the approximated electric field $E_{approx}$ is given using in equation (I) as:

$\begin{array}{c}{E}_{approx}=\frac{1}{4蟺{蔚}_o}\frac{2qd}{{\left(5d\right)}^3}\\ =\frac{2}{125}\frac{q}{4蟺{蔚}_o{d}^2}.\end{array}$

鈥︹�︹�︹�︹�︹�︹�︹�︹�︹��.. (III)

Thus, the required ratio can be given by dividing equation (III) by equation (II) as:

$\begin{array}{c}{E}_{approx}/E_{act}=\frac{\left(\frac{2}{125}\frac{q}{4蟺{蔚}_o{d}^2}\right)}{\left(\frac{160}{9801}\frac{q}{4蟺{蔚}_o{d}^2}\right)}\\ =0.9801\\ 鈮�0.98\end{array}$

Hence, the value of the ratio is$0.98$