Respuesta :
Answer:
[tex]\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }[/tex]
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
[tex]=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }[/tex]
Where
- q is the charge
- r is the distance
- [tex]E[/tex] 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[tex]=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }[/tex]
F2[tex]=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }[/tex]
⇒
F1+F2=[tex]\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }[/tex]
