Answer : [tex]\underset{E_{R}}{\rightarrow} =-2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}[/tex]
Explanation :
Given that,
Charge of particle 1 = [tex]-1.50\times10^{-7} c[/tex]
Distance x = 6 cm
Charge of particle 2 = [tex]1.50\times10^{-7} c[/tex]
Distance x = 27 cm
Total distance = [tex]\dfrac{6+27}{2}[/tex]
[tex]r = 16.5\ cm[/tex]
Particle 1 is at (6,0) and particle 2 is at (27,0) .
Therefore, midway (16.5, 0)
Now, [tex]r = \dfrac{|6-16.5|}{2} = \dfrac{|27-16.5|}{2} = 10.5\ cm[/tex]
Formula of electric field
[tex]E = \dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{r^{2}}[/tex]
Now, the the electric field due to particle 1
[tex]\underset{E}{\rightarrow}\ = -\dfrac{9\times10^{9}\times1.50\times10^{-7 }}{10.5}\ \widehat{i} \dfrac{N}{C}[/tex]
[tex]\underset{E}{\rightarrow} = \dfrac{13.5\times10^{2}}{(10.5\times10^{-2})^{2}}\widehat{i} \dfrac{N}{C}[/tex]
[tex]\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}[/tex]
Similarly, the electric field due to particle 2
[tex]\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}[/tex]
Resultant Electric field
[tex]\underset{E_{R}}{\rightarrow} = \underset{E_{1}}{\rightarrow} + \underset{E_{2}}{\rightarrow}[/tex]
[tex]\underset{E_{R}}{\rightarrow} = -2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}[/tex]
Hence, this is the required answer.
