An electric dipole is formed from two charges, ±q, spaced 0.700 cm apart. the dipole is at the origin, oriented along the y-axis. the electric field strength at the point (x, y)=(0cm,10cm) is 330 n/c . what is the charge q? give your answer in nc what is the electric field strength at the point (x, y) = (10 cm, 0 cm) n/c = ?

a)  q = 0.2 C  and b)  E = 138.37 10⁹ N/C

Explanation:

The problem let's start by defining a coordinate system, where the x-axis is horizontal and the vertical axis, the zero of the system is at the midpoint of the dipole charges.  

a) Let's calculate the field for each charge at the point y = 10 cm, x=0 or in the dipole line, for that we calculate the field that creates each charge and add them vectorially  

Charge1  

   y₁ = (0.1 - 0.035) = 0.065 m  

   E1 = k q / y₁²  

   E1 = 8.99 10 9 q / 0.065²  

  E1 = 2127.8 q  

charge 2  

  y₂ = (0.1 - (-0.035) = 0.135 m  

  E2 = k (-q) / y₂²  

  E2 = 8.99 109 q / 0.135²  

  E2 = - 493.3 q  

In an electric dipole the value of the two charges is of equal magnitude, but otherwise, suppose the positive charge is the highest  

   Et = E1 -E2  

    Et = 2127.8q -493.3q  

   330 = q (2127.8 -493.3)  

   q = 330 / 1634.5  

   q = 0.2 C

 Now suppose the negative charge is higher  

   Et = (-2127.8 +493.3) q  

   330 = -1634.5 q  

   q = -0.2 C  

b) calculate the field for the point x = 0.1 m y = 0 m  

This is a point perpendicular to the midpoint of the dipole, so that we can use the dipole equation  

    E = k 2qa / √(x² + a²)³  

    E = 8.99 10⁹ 2 0.2 0.07 / √(0.1² + 0.07²)³  

    E = 0.2517 10⁹ / (0.001819)  

     E = 138.37 10⁹ N/C

d) physical means can be used to separate the components of air.