91Ó°ÊÓ

a) Charges of the point charges, $q_1=30×{10}^{−9}\text{C}$,$q_2=40×{10}^{−9}\text{C}$

b) Distance of charges from origin$r_1=0\text{″¾}$, $r_2=0.72\text{″¾}$

c) Charge of third particle,$q_3=42×{10}^{−6}\text{ C}$

d) Distance of the third particle from origin, $r_3=0.28\text{″¾}$

e) Magnitude of acceleration of the third particle, $a=100{\text{ k³¾/²õ}}^{\text{2}}={10}^5{\text{″¾/s}}^{\text{2}}$

In presence of a large number of charges, the resultant coulomb force acting on a particle is calculated by taking the vector sum of all the forces acting on the particle due to other charges individually.

The electrostatic force of attraction or repulsion on a body by another charged body according to Coulomb’s law, ${\mathbf{F}}_{\mathrm{ij}}\mathbf{=}\frac{\mathbf{k}\mathbf{\left|}{\mathbf{q}}_i\mathbf{\right|}\mathbf{\left|}{\mathbf{q}}_j\mathbf{\right|}}{{\mathbf{\left|}{\mathbf{r}}_{\mathrm{ij}}\mathbf{\right|}}^2}$(i)

The force of a body due to Newton’s second law, $\mathbf{F}\mathbf{=}\mathbf{ma}$ (ii)

Here,${\mathbf{q}}_{\mathbf{i}\text{ }}\mathbf{and}\text{ }{\mathbf{q}}_j$ are the charged applying force on each other and ${\mathrm{r}}_{\mathrm{ij}}$ is the distance between the charges ${\mathbf{q}}_{\mathbf{i}\text{ }}\mathbf{and}\text{ }{\mathbf{q}}_j$.

Using the concept and equation (i) with the given data, the magnitude of net force on particle 3 due to particles 1 and 2 can be calculated as follows:

$\begin{array}{c}{F}_3=F_{31}+F_{32}\\ =\frac{k\left|q_1\right|\left|q_3\right|}{{\left|r_{13}\right|}^2}+\frac{k\left|q_2\right|\left|q_3\right|}{{\left|r_{23}\right|}^2}\\ =\left(\begin{array}{l}\frac{(9×{10}^{9\text{ }}{\text{N.m}}^2{\text{/C}}^2)(30×{10}^{−9}\text{ C})(42×{10}^{−6}\text{ C})}{{(0.28−0)}^2{\text{m}}^2}+\\ \frac{(9×{10}^9{\text{ N.³¾}}^2{\text{/C}}^2)(40×{10}^{−9}\text{ C})(42×{10}^{−6}\text{ C})}{{(0.28−0.72)}^2{\text{m}}^2}\end{array}\right)\\ =0.22\text{ N}\end{array}$

Now, the mass of the particle can be calculated using the above data in equation (ii) as follows:

$\begin{array}{c}m=\frac{F_3}{a}\\ =\frac{0.22\text{ N}}{{\text{10}}^5{\text{″¾/s}}^2}\\ =2.2×{10}^{−6}\text{ k²µ}\end{array}$

Hence, the value of the mass is $2.2×{10}^{−6}\text{ k²µ}$.