Eₓ = [(qx) / [(4πϵ₀r)√(x² + y² + z²)³]
Further explanation
In this problem, we will solve the derivatives of the composite function. The formula used is the chain rule as follows.
[tex]\boxed{ \ If \ f(x) = h(g(x)) \ then \ f'(x) = h'(g(x)) \times g'(x) \ }[/tex]
In words: differentiate the outer function, then multiply it by the derivative of the innermost function.
Given:
[tex]\boxed{ \ V = \frac{q}{4 \pi \epsilon_0 r} \rightarrow V = \frac{q}{4 \pi \epsilon_0 r \sqrt{x^2 + y^2 + z^2}} \ }[/tex]
Question:
Calculate the x-component of electric field by using [tex]\boxed{ \ E_x = - \frac{\delta V}{\delta x} \ }[/tex]
The Process:
This case is a derivative application for electrostatics in physics.
The potential due to a point charge q at the origin may be written as
[tex]\boxed{ \ V = \frac{q}{4 \pi \epsilon_0 r \sqrt{x^2 + y^2 + z^2}} \rightarrow V = \frac{q}{4 \pi \epsilon_0 r}(x^2 + y^2 + z^2)^{-\frac{1}{2}} \ }[/tex]
Let's determine the derivative and run the composite function rule.
Let us calculate the x-component of electric field by using
[tex]\boxed{ \ E_x = - \frac{\delta V}{\delta x} \ }[/tex]
The composite function rule tells us that
[tex]\boxed{ \ V(x) = h(g(x)) \rightarrow V'(x) = h'(g(x)) \times g'(x) \ }[/tex]
Therefore, [tex]\boxed{ \ E_x = - V'(x) \rightarrow -[h'(g(x)) \times g'(x)] \ }[/tex]
[tex]\boxed{ \ E_x = -[-\frac{1}{2} \frac{q}{4 \pi \epsilon_0 r}(x^2 + y^2 + z^2)^{-\frac{3}{2}} \times 2x] \ }[/tex]
Thus, the result is [tex]\boxed{\boxed{ \ E_x = \frac{qx}{4 \pi \epsilon_0 r}(x^2 + y^2 + z^2)^{-\frac{3}{2}} \ }}[/tex]
[tex]\boxed{\boxed{ \ E_x = \frac{qx}{4 \pi \epsilon_0 r \sqrt{(x^2 + y^2 + z^2)^3} }} \ }}[/tex]
Keywords: calculus, differential, the composite function rule, the potential due to a point charge q, at the origin, physics, application, the chain rule
