Question

Two point charges, 2.5 x 10 -6 C and -5.0 x 10 -6 C , a replaced 3.0 m apart as shown below in Fig. 1. What is the magnitude of the electric field at point P, midway between the two charges?

Answer 1

Answer:

Explanation:

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Answer 2

Answer:

The magnitude of the electric field at point P is $3\times 10^{4}N/C$.

Explanation:

Given two point charges, $q_1= 2.5 \times 10^ {-6} C$ and $q_2=-5.0 \times 10 ^{-6} C$

The distance between the charges, $r=3.0 m$

If point P, midway between the two charges, then the distance between the $q_1$ & P and $q_2$ & P is same and hence, $r=\frac{3}{2} =1.5m$

The magnitude of the electric field at at distance r by a charge q  is

$E=k\dfrac{q}{r^{2} }$

where k is the Coulomb's constant, $=9\times 10^{9} Nm^{2} /C^{2}$

Substituting the values for each charge,

the electric field due to charge $q_1$ is

$E_1=\dfrac{9\times 10^{9}\times2.5 \times 10^ {-6} }{(1.5)^{2} }=\dfrac{9\times 10^{3}\times2.5 }{2.25 }=1\times 10^{4}N/C$

the electric field due to charge $q_2$ is

$E_1=\dfrac{9\times 10^{9}\times(-5 )\times 10^ {-6} }{(1.5)^{2} }=-\dfrac{9\times 10^{3}\times5 }{2.25 }=-2\times 10^{4}N/C$

Neglecting the negative sign as it is only the magnitude,

the magnitude of electric field at point P due to both charges is

$E=E_1+E_2$

$=1\times 10^{4}+2\times 10^{4}=3\times 10^{4}N/C$

Therefore, the magnitude of the electric field at point P, midway between the two charges is $3\times 10^{4}N/C$.