Question

A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm , y = 3.2 nm .
Part A) Find the force on the ion of charge -2e. Express your answer in vector components, separated by commas, using two significant figures.

Answer 1

Answer:

$\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=<-4.30\times 10^{-11}N\,,\,-3.62\times 10^{-11}N>$

Explanation:

Given that

$Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m$

As both charges are negative so there exist force of repulsion in direction as shown in figure.

$F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N$

Angle at which force F12 is acting is

$\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o$

$F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N$

$\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=<4.30\times 10^{-11}N\,,\,3.62\times 10^{-11}N>$

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

$F_{21}=-5.63\times 10^{-11}N$

$\vec{F}_{21}=<-4.30\times 10^{-11}N\,,\,-3.62\times 10^{-11}N>$

Answer 2

The vector form of force on the ion of the charge is  $(4.30 \times 10^{-11}) i+ (3.62 \times 10^{-11})j$.

Electrostatic Force:

The force exerted by any charged entities on the other charged particles is known as the Electrostatic force.

Given data:

The magnitude of the charge on a negative ion is, -2e.

The magnitude of the charge on the second negative ion is, -3e.

The corresponding distance is,  $x =3.8 \;\rm nm = 3.8 \times 10^{-9}\;\rm m$  and    $y =3.2 \;\rm nm = 3.2 \times 10^{-9}\;\rm m$.

The net distance is,

$r = \sqrt{x^{2}+y^{2}}\\\\ r = \sqrt{(3.8 \times 10^{-9})^{2}+(3.2 \times 10^{-9})^{2}}\\\\ r = 4.96 \times 10^{-9} \;\rm m$

As both charges are negative, there exists a force of repulsion. Then the magnitude of the force is,

$F = \dfrac{k \times (-2e) \times (-3e)}{r^{2}}\\\\\\ F = \dfrac{9 \times 10^{9} \times (-2 \times 1.6 \times 10^{-19}) \times (-3 \times 1.6 \times 10^{-19})}{(4.96 \times 10^{-9})^{2}}\\\\ F = 5.63 \times 10^{-11} \;\rm N$

The angle at which force F is acting is calculated as,

$\theta = tan^{-1} \dfrac{x}{y}\\\\\\ \theta = tan^{-1} \dfrac{3.8}{3.2}\\\\\\ \theta = 40.1^{\circ}$

Now, the x-component of the above force is,

$F_{x}=F \times cos \theta\\\\ F_{x}=(5.63 \times 10^{-11}) \times cos(40.1)\\\\ F_{x}=4.306 \times 10^{-11} \;\rm N$

And the y-component of the given force is,

$F_{y}=Fsin \theta\\\\ F_{y}=(5.63 \times 10^{-11}) \times sin(40.1)\\\\ F_{y}=3.62 \times 10^{-11}\;\rm N$

Then the vector form of the given force is,

$F = F_{x}+F_{y}\\\\ F = (4.30 \times 10^{-11}) i+ (3.62 \times 10^{-11})j$

Thus, we can conclude that the vector form of force on the ion of the charge is  $(4.30 \times 10^{-11}) i+ (3.62 \times 10^{-11})j$.

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