91Ó°ÊÓ

a) Charges of the four particles,$q_1=−3.20×{10}^{−19}\text{C}$,$q_2=+3.20×{10}^{−19}\text{C}$,$q_2=+3.20×{10}^{−19}\text{C}$and $q_4=+3.20×{10}^{−19}\text{C}$.

b) Angle made by the displacement vector of particle 1 with the negative x-axis,$\mathrm{θ}=35\text{ }°\text{C}$ .

c) Distance between particle 4 and particle 1,$d_1=3\text{ c³¾}$

d) Distance between particles 2 or 3 and 4,$d_2=d_3=2\text{ c³¾}$

When more than one force is acting on a body, we follow the superposition principle to calculate the net force acting on the given charge or body. The superposition principle states that the vector sum of all the forces acting on the body due to other charged bodies present at a distance from it will give the net force acting on the body.

The electrostatic force of attraction or repulsion on a body by another charged body according to Coulomb’s law in unit vector notation, ${\stackrel{→}{F}}_{ij}=\frac{k\left|q_i\right|\left|q_j\right|}{{\left|r_{ij}\right|}^2}{\widehat{r}}_{ij}$ (i)

The unit-position vector making an angle,${\widehat{r}}_{ij}=\cos \mathrm{θ}\widehat{i}+\sin \mathrm{θ}\widehat{j}$ (ii)

The magnitude of a displacement vector,$r=\sqrt{x^2+y^2}$ (iii)

The direction of a vector, $\mathrm{θ}={\tan }^{−1}\left(\frac{y−component}{x−component}\right)$ (iv)

Here, x represents the magnitude of component of displacement vector$\left(\stackrel{→}{r}\right)$ along x-axis. Similarly, y represents the magnitude component along y-axis.

Using equations (i), the force acting on particle 4 by particle i can be given as follows:

${\stackrel{→}{F}}_{4i}=\frac{k\left|q_4\right|\left|q_i\right|}{{\left|r_{4i}\right|}^2}{\widehat{r}}_{4i}\mathrm{.........................}\left(a\right)$

For particle 1:

As the charge on particle 1 is opposite to that on particle 4, the force is attractive thus, the force on particle 4 due to particle 1 can be given using equation (a) and (ii) as:

$\begin{array}{c}\stackrel{→}{F_{41}}=\frac{(9×{10}^9{\text{ N.³¾}}^2{\text{/C}}^2)|3.20×{10}^{−19}\text{ C}||3.20×{10}^{−19}\text{ C}|}{{\left|0.03\text{″¾}\right|}^2}(−\cos (35°)\widehat{i}−\sin (35°)\widehat{j})\\ =(1.02×{10}^{−24\text{ }}\text{N})(−0.82\text{ }\widehat{i}−0.57\text{ }\widehat{j})\\ =−(8.39×{10}^{−25}\text{ N})\hat{i}−(5.87×{10}^{−25}\text{ N})\hat{j}\end{array}$

Similarly, the force acting on particle 4 due to particle 2 can be given using equation (a) and (ii) as follows: (since it a repulsive force)

$\begin{array}{c}\stackrel{→}{F_{42}}=\frac{(9×{10}^9{\text{ N.³¾}}^2{\text{/C}}^2)|3.20×{10}^{−19}\text{ C}||3.20×{10}^{−19\text{ }}\text{C}|}{{\left|0.02\text{″¾}\right|}^2}(−\cos \left({90}^0\right)\widehat{i}−\sin \left({90}^0\right)\widehat{j})\\ =(2.30×{10}^{−24}\text{ N})(−\widehat{j})\\ =−(2.30×{10}^{−24}\text{ N})\widehat{j}\end{array}$

Similarly, the force acting on particle 4 due to particle 3 can be given using equation (a) and (ii) as follows: (since it a repulsive force)

$\begin{array}{c}\stackrel{→}{F_{43}}=\frac{(9×{10}^9{\text{ N.³¾}}^2{\text{/C}}^2)|6.40×{10}^{−19}\text{ C}||3.20×{10}^{−19}\text{ C}|}{{\left|0.02\text{″¾}\right|}^2}(−\cos (0°)\widehat{i}−\sin (0°)\widehat{j})\\ =(4.61×{10}^{−24}\text{ N})(−\widehat{i})\\ =−(4.61×{10}^{−24}\text{ N})\widehat{i}\end{array}$

Thus, the net force acting on particle 4 due to all the three particles can be given as follows:

$\begin{array}{c}\stackrel{→}{F_4}=\stackrel{→}{F_{41}}+\stackrel{→}{F_{42}}+\stackrel{→}{F_{43}}\\ =−(8.39×{10}^{−25}\text{ N})\widehat{i}−(5.87×{10}^{−25}\text{ N})\widehat{j}−(2.30×{10}^{−24}\text{ N})\widehat{j}−(4.61×{10}^{−24}\text{ N})\widehat{i}\\ =−(5.44×{10}^{−24}\text{ N})\widehat{i}−(2.89×{10}^{−24}\text{ N})\widehat{j}\end{array}$

Now, the magnitude of the force can be given using equation (iii) as follows:

$\begin{array}{c}\left|\stackrel{→}{F_4}\right|=\sqrt{{(−5.44×{10}^{−24}\text{ N})}^2+{(−2.89×{10}^{−24}\text{ N})}^2}\\ =6.16×{10}^{−24}\text{ N}\end{array}$

Hence, the magnitude of the resultant force is.$6.16×{10}^{−24}\text{ N}$

Using the net force value from part (a) calculations in equation (ii), the direction of the net force acting on particle 4 due to all three particles can be given as follows:

$\begin{array}{c}\mathrm{θ}={\tan }^{−1}\left(\frac{−(2.89×{10}^{−24}\text{ N})}{−(5.44×{10}^{−24}\text{ N})}\right)\\ =208°,\text{measuredcounter-clockwisefromx-axis}\end{array}$

Hence, the direction of the force is.$208°,\text{measuredcounter-clockwisefromx-axis}$