Question

Electric charges + 1000 C are placed at points A and
8. respectively at a distance of 2 m from each other.
Calculate the electric fields ar (1) mid-point of the line
AB and (D) at a point at equal distances of 4 m from
cach charge. Ans. () 1.8 10'NC. along AB :​

Answer 1

Answer:

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 E2­. 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

2nd 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.