Answer:
The net electric field at the midpoint is 6.85 x 10^7 N/C.
Explanation:
q = − 8.3 μC
q' = + 7.8 μC
d = 9.2 cm
d/2 = 4.6 cm
The electric field due to the charge q at midpoint is
leftwards
The electric field due to the charge q' at midpoint is

The resultant electric field at mid point is
E'' = E + E' = (3.53 + 3.32) x 10^7 = 6.85 x 10^7 N/C
Complete Question
The complete question is shown on the first uploaded image
Answer:
The magnetic field is 
And the direction is 
Explanation:
From the question we are told that
The magnetic field at the center is 
Generally magnetic field is mathematically represented as

We are told that it is equal to 1mT
So

From the first diagram we see that the effect of the current flowing in the circular loop is (i.e the magnetic field generated)

This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)

and for the larger semi-circular loop is

Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction
So the net magnetic field would be





Recall 
So

Using the Right-hand rule we see that the direction is into the page which is 
Answer:
Book of mass 2 kg is lifted from floor to the table. The height between floor and the table is 1.5 m. Calculate the work done by gravitational force.
Answer:
distance is not a vector. it is scalar
Answer:
a

b

Explanation:
From the question we are told that
The wavelength of the light is 
The distance of the slit separation is 
Generally the condition for two slit interference is

Where m is the order which is given from the question as m = 2
=> ![\theta = sin ^{-1} [\frac{m \lambda}{d} ]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20sin%20%5E%7B-1%7D%20%5B%5Cfrac%7Bm%20%5Clambda%7D%7Bd%7D%20%5D)
substituting values

Now on the second question
The distance of separation of the slit is

The intensity at the the angular position in part "a" is mathematically evaluated as
![I = I_o [\frac{sin \beta}{\beta} ]^2](https://tex.z-dn.net/?f=I%20%20%3D%20%20I_o%20%20%5B%5Cfrac%7Bsin%20%5Cbeta%7D%7B%5Cbeta%7D%20%5D%5E2)
Where
is mathematically evaluated as

substituting values


So the intensity is
![I = I_o [\frac{sin (0.06581)}{0.06581} ]^2](https://tex.z-dn.net/?f=I%20%20%3D%20%20I_o%20%20%5B%5Cfrac%7Bsin%20%280.06581%29%7D%7B0.06581%7D%20%5D%5E2)
