Question

The potential due to a point charge q at the origin may be written as v=q4πϵ0r=q4πϵ0x2+y2+z2√ part a calculate ex using equation ex=−∂v∂x.

Answer 1

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.

$\boxed{ \ If \ f(x) = h(g(x)) \ then \ f'(x) = h'(g(x)) \times g'(x) \ }$

In words: differentiate the outer function, then multiply it by the derivative of the innermost function.

Given:

$\boxed{ \ V = \frac{q}{4 \pi \epsilon_0 r} \rightarrow V = \frac{q}{4 \pi \epsilon_0 r \sqrt{x^2 + y^2 + z^2}} \ }$

Question:

Calculate the x-component of electric field by using $\boxed{ \ E_x = - \frac{\delta V}{\delta x} \ }$

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

$\boxed{ \ V = \frac{q}{4 \pi \epsilon_0 r \sqrt{x^2 + y^2 + z^2}} \rightarrow V = \frac{q}{4 \pi \epsilon_0 r}(x^2 + y^2 + z^2)^{-\frac{1}{2}} \ }$

• The outside function is $\boxed{ \ h(x) = \frac{q}{4 \pi \epsilon_0 r}(g(x))^{-\frac{1}{2}} \ }$
• The inside function is $\boxed{ \ g(x) = x^2 + y^2 + z^2 \ }$

Let's determine the derivative and run the composite function rule.

• The outside function: $\boxed{ \ h'(x) = -\frac{1}{2} \frac{q}{4 \pi \epsilon_0 r}(g(x))^{-\frac{3}{2}} \ }$
• The inside function: $\boxed{ \ g'(x) = 2x \ }$

Let us calculate the x-component of electric field by using

$\boxed{ \ E_x = - \frac{\delta V}{\delta x} \ }$

The composite  function rule tells us that

$\boxed{ \ V(x) = h(g(x)) \rightarrow V'(x) = h'(g(x)) \times g'(x) \ }$

Therefore, $\boxed{ \ E_x = - V'(x) \rightarrow -[h'(g(x)) \times g'(x)] \ }$

$\boxed{ \ E_x = -[-\frac{1}{2} \frac{q}{4 \pi \epsilon_0 r}(x^2 + y^2 + z^2)^{-\frac{3}{2}} \times 2x] \ }$

Thus, the result is $\boxed{\boxed{ \ E_x = \frac{qx}{4 \pi \epsilon_0 r}(x^2 + y^2 + z^2)^{-\frac{3}{2}} \ }}$

$\boxed{\boxed{ \ E_x = \frac{qx}{4 \pi \epsilon_0 r \sqrt{(x^2 + y^2 + z^2)^3} }} \ }}$

Learn more

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Keywords: calculus, differential, the composite function rule, the potential due to a point charge q, at the origin, physics, application, the chain rule

Answer 2

The electric potential due to a point charge q at the origin is given by
$v= \frac{q}{4 \pi \epsilon _{0} r} = \frac{q}{4 \pi \epsilon _{0} \sqrt{x^{2}+y^{2}+z^{2}} }$

The x-component of electric field is
$e_{x} = - \frac{\partial v}{\partial x} = -\frac{q}{4 \pi \epsilon {0}}(- \frac{1}{2}) \frac{2x}{(x^{2}+y^{2}+z^{2})^{3/2}} = \frac{qx}{4 \pi \epsilon _{0} (x^{2}+y^{2}+z^{2})^{3/2}}$

Answer:
$e_{x} = \frac{qx}{4 \pi \epsilon _{0} (x^{2}+y^{2}+z^{2})^{3/2}}$