Question

calculate the electric field intensity at a point distance 10m from a point charge of 2 coulomb in air​

Answer 1

Answer:

Explanation:

E=14π∈0qr2 , where ∈0 is absoulute electrical permittively of free space.

Answer 2

Answer:

 Finding closed surface for application of Gauss Law

The square can be considered as one face of a cube of edge 10cm with a centre where charge q is placed. 

According to Gauss’s law for a cube, total electric flux is through all its six faces.

ϕtotal=∈0q

Step 2: Flux through one face or square

As the charge is placed symmetrically to each face of the cube, thus electric flux passing through each face is equal.

So, Electric flux through one face of the cube i.e., through the square,

ϕ1=6ϕtotal=61∈0q

⇒  ϕ1=6×8.854×10−12N−1m−2C210×10−6C=1.88×105Nm2C−1

Therefore, electric flux through the square is 1.88×105Nm2C−1.

Explanation:

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