Respuesta :

Answer:

[tex]y = \frac{d}{\sqrt2}[/tex]

Explanation:

As we know that electric field due to each charge at any given height y due to the point charges is given by

[tex]E = \frac{kq}{d^2 + y^2}[/tex]

now we know that horizontal component of electric field due to each charge will cancel out while vertical component will be added due to both

So here we can say

[tex]E_{net} = \frac{kqy}{(d^2 + y^2)^{3/2}}[/tex]

now we know that for maximum value of Electric field

[tex]\frac{dE}{dy} = 0[/tex]

after solving this we will have

[tex]y = \frac{d}{\sqrt2}[/tex]

The electric field at y-axis will be maximum at; y = d/√2

Electric field

Let the net field intensity at P be E⁰.

Thus;

E⁰ = 2 × E × cos θ

We know that formula for electric field is;

E = kq/r²

Thus;

E⁰ = 2 × (kq/r²) × cos θ

But from the given image, we can say that;

cos θ = y/r and r = √(d² + y²)

Thus, our net field intensity equation is now;

E⁰ = 2 × (kq/(√(d² + y²))²) × y/√(d² + y²)

Simplifying that gives us;

 [tex]E^{0} = \frac{2kqy}{(\sqrt{d^{2} +y^{2}})^{\frac{3}{2}}} [/tex]

When we differentiate that with respect to y from online differentiation calculator and equate to zero, we arrive at;

d² + y² = 3y²

Thus;

3y² - y² = d²

2y² = d²

Taking square root of both sides gives;

y = d/√2

Read more about Electric Field at; https://brainly.com/question/1592046

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