Answer:
![\frac{1}{4*(pie)*\text{E}} *(q)/(I^(2) )+\frac{1}{4*(pie)*\text{E}} *(-q)/(I^(2) )](https://img.qammunity.org/2020/formulas/physics/middle-school/pytswid0rw7cswmmiu8il3vgxwpis1epla.png)
Step-by-step explanation:
As given point p is equidistant from both the charges
It must be in the middle of both the charges
Assuming all 3 points lie on the same line
Electric Field due a charge q at a point ,distance r away
![=\frac{1}{4*(pie)*\text{E}} *(q)/(r^(2) )](https://img.qammunity.org/2020/formulas/physics/middle-school/3h9xj2zfvtzb96fzxji2gu25mt6pic0nnk.png)
Where
- q is the charge
- r is the distance
-
is the permittivity of medium
Let electric field due to charge q be F1 and -q be F2
I is the distance of P from q and also from charge -q
⇒
F1
![=\frac{1}{4*(pie)*\text{E}} *(q)/(I^(2) )](https://img.qammunity.org/2020/formulas/physics/middle-school/pwbgnih8x18n4ki1gkg3x3dezs7v0gzubt.png)
F2
![=\frac{1}{4*(pie)*\text{E}} *(-q)/(I^(2) )](https://img.qammunity.org/2020/formulas/physics/middle-school/hxdoa3uctgrj7lqsv6wrlyw5pw1h0kw7a7.png)
⇒
F1+F2=
![\frac{1}{4*(pie)*\text{E}} *(q)/(I^(2) )+\frac{1}{4*(pie)*\text{E}} *(-q)/(I^(2) )](https://img.qammunity.org/2020/formulas/physics/middle-school/pytswid0rw7cswmmiu8il3vgxwpis1epla.png)