Question

Please also include formulas used

2. Two positive charges are located a distance of 0.58 m apart on the x axis. At the origin, the charge is 5.5 x 104c, and at x-0.58m the charge is 3.3 x 10-c. At what point is the electric field equal to zero

Answer 1

As the two charges are of same sign, so, the electric field is zero in between the charges. So, we assume that the zero electric field point to be at a distance x from the origin. And if we assume that the distance between the two charges is d, then, we get
  
As the two charges are of same sign,. So the electric field is zero in between the charges So, we assume that zero electric field point tio be at a distance x from the origin. And if we assume that the distance between the two charges is d, then we get E(x) = 1/4 pi elementof_0 [q_1/x^2 - q_2/(d - x)^2] where q_1 is situated at the origin, And q_2 is situates at the distance d So, for E(x) = 0 we require E(x) = 1/4 pi elementof_0 [q_1/x^2 - q_2/(d - x)^2] = 0 implies q_1/x^2 - q_2/(d - x)^2 implies q_2 x^2 = q_1(d - r)62 Now for the given values q_1 = 5.5 times 10^-8 C, q_2 = 3.3 times 10^-8 C And so 3.3 times 10^-8 x^2 = 5.5 times 10^-8 (d - x)^2 implies 3x^2 = 5(d-x)^2 implies x^2 -= 5/3(d-x)^2 implies x = plusminus squareroot 5/3 (d -x).

where, $q_1$ is situated at the origin. And $q_2$ is situated at the distance d.
So, for E(x) = 0, we require
  

  
Now for the given values,
     
and so,
  
   
   
  
  
   For the +ve sign, we have
   
   
   
   
And for d = 0.58 m, we have
  
  

And for -ve sign , we have
     
   
   
   
And for d = 0.58 m, we have
  
  
And as this is greater than 0.58 m, so, this can't be the answer. So, the correct answer is