• Electric charges -+1000 microcoulomb are placed at points A and B respectively at a distance of 2m from each other.Calculate the electric feilds at (a)midpoint of the line AB and (b)at a point at equal distances of 4m from each charge(ans (a)1.8*10 N/C along AB ..(b)2.8 *10 to the power 5 N/C,parallel to AB)

Since electric field is a vector quantity, we will not consider the signs of the charges in this question.

We know that electric field is given as:

E = kq/r²

Here k is the constant = 9*10⁹

q is the charge = 1000 *10^-6C

r is the distance of the charge from mid point = 1m

let electric field due to 1 charge be E₁ and due to the other be E_2­. Since both the charges are equal in magnitude, E₁ = E₂. Also, the final E will be E₁ + E₂

hence 2kq/r²,

2*9*10⁹ *10³ *10^-6

1.8 *10^-5N/C

2^nd case

Here the point at which the electric field has to be found forms an isosceles triangle with both the charges each 4m from the equatorial point .

This point lies on the equatorial point.

E = k*2aq/(r² +a²)³

Here r = altitude = √15

a = 1m = ½ the distance from each charge.

K = 9*10⁹

2a = 2m

Substituting the values in the above formula we get,

E = 2.8 *10⁵N/C.