Question

two point charges each of 20uc are placed 50cm apart in air .what is the electric field intensity at the midpoint on the line joining the centre of two point charges​

Answer 1

Answer:

7.2×10*5

Explanation:

Kq÷r*2

= 9×10*9×20×10*-6÷(50×50×10*-4)

Answer 2

Answer:

Explanation:

Two point charges are placed at two end of a line,

Charge on each charge = $q_1=q_2=20*10^{-6} C$

Distance by which they are separated= 50cm=0.5m

We have to find the electric field at the center which is 0.25m away from charge.

From the figure we can see that$E_1=\frac{kq}{(r/2)^2}=\frac{9*10^9*20*10^{-6}}{0.25^2}= 2.88*10^6 NC$

$E_2$ is also same but opposite in direction ⇒$E_2=-2.88*10^6 NC$

Hence the net electric field at the center is =$E_{net}=E_1+E_2=0$