Definition of electric field dE due to a small charge dQ

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SUMMARY

The equation for the electric field due to a small charge, expressed as dE = (k dQ / r²)(cosθ i + sinθ j), is derived from Coulomb's law. A small charge dQ can be treated as a point charge, allowing the use of this formula to calculate the electric field at a distance r. The components of the electric field in Cartesian coordinates are represented by the unit vector r̂, which incorporates the trigonometric functions cosine and sine to account for the angle θ. Understanding this derivation is essential for applying the equation correctly in physics problems.

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  • Coulomb's Law
  • Vector calculus
  • Understanding of unit vectors
  • Basic trigonometry
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twotaileddemon
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Homework Statement



Hi...
How can one be sure that
[tex] d\textbf{E}=\frac{kdQ}{r^2}(\cos\theta \textbf{i} + \sin\theta \textbf{j})[/tex]
Is true for a small charge dQ
Is there some derivation for this?

Homework Equations

The Attempt at a Solution



I have a problem for homework that uses this equation... I know how to solve the problem, but I feel like I should know how this equation works since I don't want to just mindlessly plug in information... Thanks.
 
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A small enough charge dQ can be considered a point charge. The equation is just a statement of Coulomb's law.

[tex] d\textbf{E}=\frac{kdQ}{r^2}\hat{r}[/tex]

The unit vector [tex]\hat{r}[/tex] in Cartesian coordinates gives you what you have.
 
naresh said:
A small enough charge dQ can be considered a point charge. The equation is just a statement of Coulomb's law.

[tex] d\textbf{E}=\frac{kdQ}{r^2}\hat{r}[/tex]

The unit vector [tex]\hat{r}[/tex] in Cartesian coordinates gives you what you have.

Ohh okay I get it now, thanks. I understand where the sin and cosine come from now
:)
Thanks so much!
 

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