91Ó°ÊÓ

1. Value of the identical charges,${\mathrm{q}}_1={\mathrm{q}}_2=50×{10}^{-6}\mathrm{C}$
2. The charges are fixed at $\mathrm{x}=Â\pm 3\mathrm{m}$.
3. A particle of charge$\mathrm{q}=-15\mathrm{μ°ä}$ is released from rest from positive y-axis.
4. Symmetry causes the particle to move along the y-axis.
5. The kinetic energy of the particle$\mathrm{K}=1.2\mathrm{J}$ at point is$\mathrm{x}=0,\mathrm{y}=4.0$m

Using the concept of energy, solve for the potential energy of the particle at the given point and the origin. Now, using these values in the conservation law of energy, obtain the required value of the speed of the particle. Now, using a similar concept, obtain the required coordinate of the particle along the y-axis where the particle rests momentarily.

From the law of conservation of the energy, the sum of the initial potential energy (U_i) and initial kinetic energy (K_i) is equal to the sum of the final potential energy (U_f) and the final kinetic energy (K_f).

${\mathrm{U}}_{\mathrm{i}}+{\mathrm{K}}_{\mathrm{i}}={\mathrm{U}}_{\mathrm{f}}+{\mathrm{K}}_{\mathrm{f}}$ …… (i)

The kinetic energy of a moving object depends on its mass and its speed. The mathematical expression to calculate the kinetic energy is,

$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$ …â¶Ä¦(¾±¾±)

Here, m is the mass of the object and v is the speed of the object.

The electrostatic potential energy of a charge at a point in an electric field is,

$\mathrm{U}=\mathrm{Vq}$ …â¶Ä¦(¾±¾±¾±)

Here, k is the Coulomb’s constant and q is the charge.

The electric potential of a charge is as follows:

$\mathrm{V}=\frac{\mathrm{kq}}{\mathrm{r}}$ …… (iv)

The distance between the point $\left(0,4.0\mathrm{m}\right)$and the point$\left(+3.0\mathrm{m},0\mathrm{m}\right)$is,

$\begin{array}{rcl}\mathrm{r}& =& \sqrt{{\left(3.0\mathrm{m}-0\mathrm{m}\right)}^2+{\left(4.0\mathrm{m}-0\mathrm{m}\right)}^2}\\ & =& 5.0\mathrm{m}\end{array}$

By the symmetry, the distance between the point$\left(\text{0,4.0m}\right)$ and the point$\left(\text{-3.0m,0m}\right)$ is also equal to $5.0\text{m}$.

The total electric potential due to the two identical charge of $50\mathrm{μ}\text{C}$at the point $(0,4.0\text{m})$is given using equation (iv) as:

${\mathrm{V}}_1=\frac{{\mathrm{kq}}_1}{{\mathrm{r}}_1}+\frac{{\mathrm{kq}}_2}{{\mathrm{r}}_2}$

Substitute the values and solve as:

$\begin{array}{c}{\mathrm{V}}_1=\frac{(9×{10}^9\text{ }\frac{{\text{Nm}}^{\text{2}}}{{\text{C}}^{\text{2}}})(50×{10}^{−6}\text{ C})}{5.0\text{m}}+\frac{(9×{10}^9\text{ }\frac{{\text{Nm}}^{\text{2}}}{{\text{C}}^{\text{2}}})(50×{10}^{−6}\text{C})}{5.0\text{m}}\\ =1.8×{10}^5\text{ }\text{V}\end{array}$

Thus, the potential energy of the system is given using equation (iii) as given:

$\begin{array}{c}{\mathrm{U}}_{\mathrm{i}}=(1.8×{10}^5\text{V})(−15×{10}^{−6}\text{C})\\ =−2.7\text{J}\end{array}$

Potential at the origin due to the identical charges is given using equation (iv) as follows:

$\begin{array}{c}{V}_2=\frac{(9×{10}^9\text{ }\frac{{\text{Nm}}^{\text{2}}}{{\text{C}}^{\text{2}}})(50×{10}^{−6}\text{ C})}{3.0\text{m}}+\frac{(9×{10}^9\text{ }\frac{{\text{Nm}}^{\text{2}}}{{\text{C}}^{\text{2}}})(50×{10}^{−6}\text{ C})}{3.0\text{m}}\\ =3.0×{10}^5\text{ V}\end{array}$

Thus, the potential energy of the system is given using equation (iii) as follows:

$\begin{array}{c}{\mathrm{U}}_{\mathrm{f}}=(3.0×{10}^5\text{V})(−15×{10}^{−6}\text{C})\\ =−4.5\text{J}\end{array}$

Now, using energy conservation of equation (i), we can get the kinetic energy of the particle as follows:

$K_f=U_i+K_i−U_f$

Substitute the values and solve as:

$\begin{array}{c}{K}_f=(−2.7\text{J})+(1.2\text{J})−(−4.5\text{J})\\ =3.0\text{J}\end{array}$

Therefore, the final kinetic energy of the charge at the origin is $3.0\text{J}$.

Let$(0,y)$ be the final location of the$−15\mathrm{μ}\text{C}$ charge at which the charge comes to rest.

The distance between the point$(0,y)$ and the point$\left(\text{+3.0m,0m}\right)$ is given as:

$\begin{array}{c}r=\sqrt{{(3.0\text{m}−0\text{m})}^2+{(y−0\text{m})}^2}\\ =\sqrt{9+y^2}\text{m}\end{array}$

By the symmetry, the distance between the point$(0,y)$ and the point $\left(\text{-3.0m,0m}\right)$is also equal to $\sqrt{9+y^2}\text{m}$.

Now, the potential due to the two identical charge of$50\mathrm{μ}\text{C}$ at equation the point$(0,y)$ is given using equation (iv) as:

$\begin{array}{c}{V}_3=\frac{(9×{10}^9\text{ }\frac{{\text{Nm}}^{\text{2}}}{{\text{C}}^{\text{2}}})(50×{10}^{−6}\text{ C})}{\sqrt{9+y^2}\text{m}}+\frac{(9×{10}^9\text{ }\frac{{\text{Nm}}^{\text{2}}}{{\text{C}}^{\text{2}}})(50×{10}^{−6}\text{ C})}{\sqrt{9+y^2}\text{m}}\\ =\frac{9.0×{10}^5}{\sqrt{9+y^2}}\text{V}\end{array}$

Again, the potential energy of the$−15\mathrm{μ}C$ charge at the location$(0,y)$ is given using equation (iii) as:

$\begin{array}{c}{U}^{\text{'}}=\left(\frac{9.0×{10}^5}{\sqrt{9+y^2}}\text{V}\right)(−15×{10}^{−6}\text{C})\\ =−\frac{13.5}{\sqrt{9+y^2}}\text{J}\end{array}$

Now, using the given data in equation (i), we can get the coordinate at which the particle stops momentarily as:

$\begin{array}{c}−2.7\text{J}+1.2\text{J}=−\frac{13.5}{\sqrt{9+y^2}}\text{J}+0\\ 1.5\text{J}=\frac{13.5}{\sqrt{9+y^2}}\\ \sqrt{9+y^2}=\frac{13.5\text{J}}{1.5\text{J}}\\ 9+y^2={\left(\frac{13.5}{1.5}\right)}^2\end{array}$

Substitute the values and solve as:

$\begin{array}{c}y=\sqrt{{\left(\frac{13.5J}{1.5J}\right)}^2−9}\\ =−8.48\text{m}\\ =−8.5\text{m}\end{array}$

Therefore, the particle will momentarily stop at $−8.5\text{″¾}$.