Answer :
Answer: 1.653 × 10⁴ N/C
Explanation: Given Q= 3.6× 10^-6
we assume there is a charge q placed at a distance r = 1.4m exerting an electric field strength towards Q.
Hence
F = kQq/r²
K= 9× 10^9
Now recall that E = F/q (electric field due to that charge)
E =( kQq/r²)/q
E = kQ/r² substituting the values gave the answer above.
The electric field depends on the magnitude of the charge. The electric field is [tex]1.653\times 10^4 \;\rm N/C[/tex].
What is an electric field?
The electric field can be defined as an electric property associated with each point in the space where a charge is present in any form.
Given that the magnitude of point charge q is 3.6 mC and it is placed at a distance d of 1.4 m.
The electric field is given as,
[tex]E = \dfrac{q}{4 \pi \epsilon_0 d^2}[/tex]
Where [tex]\epsilon_0[/tex] is the permittivity of free space which is [tex]8.85\times 10^{-12}[/tex].
[tex]E = \dfrac { 3.6\times 10^{-6}}{4\times 3.14\times 8.85\times 10^{-12} \times 1.4^2}[/tex]
[tex]E = 1.653 \times 10^4 \;\rm N/C[/tex]
Hence we can conclude that the electric field is [tex]1.653\times 10^4 \;\rm N/C[/tex].
To know more about the electric field, follow the link given below.
https://brainly.com/question/26446532.