Question

two point charges 0.01 micro coulomb and -0.01 micro coulomb are placed 10 cm apart in vacuum calculate the magnitude of electric field intensity at the middle point of the line joining the charges and mention its direction ​

Answer 1

The magnitude of electric field intensity at the middle point of the line joining the charges is $7.2\times 10^4\ N\ C^{-1}$ and direction is towards negative charge .

Given :

Two point charges 0.01 micro coulomb and -0.01 micro coulomb are placed 10 cm apart in vacuum .

Let , us assume that right side is positive and left side is negative and positive charge is at left side .

Now , electric field at the center is given by :

$E=\dfrac{kq}{r^2}+\dfrac{kq}{r^2}\\\\E=\dfrac{2kq}{r^2}$

( because electric field due to both the charge is in right direction )

Putting all values in above equation , we get :

$E=\dfrac{2\times9\times 10^9\times0.01\times 10^{-6}}{0.05^2}\\\\E=7.2\times 10^4\ N\ C^{-1}$

Since , sign is positive therefore direction is towards right side .

Hence , this is the required solution .

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Electrostatics

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