Please show FBD and calculations for better understandingSol

Please show FBD and calculations for better understanding:

Solution

The electric field due to a point charge at a distance \'r\' from it, is given by E = Q/4*pi*epsilon*r*r

we will denot 4*pi*epsilon as K

Hence E = Q/K*r*r

a.) In the given system, we have two charges, one at x = +0.15m and other at -0.15m

Now, for x = 0.300 m,

Electric field due to both the charges is towards horizontal x axis at the given point.

hence total electric field would be the arithematic sum of the two and will be along positive x axis.

Hence Enet = Q/K[1/0.15*0.15 + 1/0.45*0.45]

E = 6 x 10^-9 x 8.9875 x 10^9 [49.38] = 2.66 x 10^3 Newtons per coulomb

b.) At the origin,the horizontal components of the fields will cancel out as they would act against each other from both the chages. Hence the net electric field at origin will be zero.