Question

3. A point charge of – 6.50 C is located at the origin. What is the magnitude and direction of the electrical field on the y-axis at 8.20 cm?

Answer 1

Solution:-

Given-

q = -6.50C   

r = 8.20 cm, it convert into meter r = 0.082 m

The strength of electric field at any point in the region is described in terms of quantity called electric intensity of electric field. It is denoted by E.

The intensity of an electric field at any point in an electric field is defined as the force acting on a unit positive charge placed at that point.

1.       To find the electric field

E = k * q1 / r²

K = Boltzmann constant and its value is 9*10⁹

K = 1/4π€0 = 9*10⁹

r = distance

E = k * q1 / r²

           = 9*10⁹ * -6.50 / 0.082

    E = - 7.13*10¹¹ N/C

Magnitude of electric field on the y axis is,     E = - 7.13*10¹¹ N/C