Question

Find electric field at point p which is a distance l away from the both +q and -q

Answer 1

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

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

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

• q is the charge
• r is the distance
• $E$ 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$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2=$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$