Answer:
[tex]0.0018\ \text{N/C}[/tex] towards the right.
[tex]0.001\ \text{N/C}[/tex] towards the right.
Explanation:
[tex]q_1=4\ \mu\text{C}[/tex]
[tex]q_2=14\ \mu\text{C}[/tex]
[tex]Q=5\ \text{nC}[/tex]
[tex]r_1=r_2=0.5\ \text{m}[/tex]
Let [tex]Q[/tex] be placed at origin so [tex]q_1[/tex] becomes negative and [tex]q_2[/tex] becomes positive
Electric field is given by
[tex]E=\dfrac{kq_1Q}{r_1^2}+\dfrac{kq_2Q}{r_2^2}\\\Rightarrow E=\dfrac{kQ}{r^2}(q_1+q_2)\\\Rightarrow E=\dfrac{9\times10^{9}\times 5\times10^{-9}}{0.5^{2}}(-4\times10^{-6}+14\times10^{-6})\\\Rightarrow E=0.0018\ \text{N/C}[/tex]
The electric field halfway between the points is [tex]0.0018\ \text{N/C}[/tex] towards the right.
[tex]r_1=0.5\ \text{m}[/tex]
[tex]r_2=1+0.5=1.5\ \text{m}[/tex]
[tex]E=9\times 10^9\times 5\times 10^{-9}(\dfrac{4\times 10^{-6}}{0.5^2}+\dfrac{14\times 10^{-6}}{1.5^2})\\\Rightarrow E=0.001\ \text{N/C}[/tex]
The electric field halfway between the points is [tex]0.001\ \text{N/C}[/tex] towards the right.
