Why is an electric force conservative?

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Why is an electric force conservative?

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Answer 1 · 2018-03-17T14:35:01

$\oint_{l} k \frac{q_{1} q_{2}}{r^{2}} ˆ r \cdot d l = 0$ $\oint_lk(q_1q_2)/r^2 \hatr* dl = 0$ ; from point $a$ $a$ to point $a$ $a$

The work done by the electric force along a path starting and ending at the same point $a$ $a$ is 0. Hence, the electric force is conservative.

Explanation:

We are given:

$F (\to r) = k \frac{q_{1} q_{2}}{r^{2}} ˆ r$ $F(\vec r) = k(q_1q_2)/(r^2)\hatr$

Force is conservative if work done along a path that starts and ends at the same point is 0.

Work is given as:

$\oint_{l} F (\to r) \cdot d l = \oint_{l} k \frac{q_{1} q_{2}}{r^{2}} ˆ r \cdot d l$ $\oint_l F(\vec r) * dl = \oint_l k(q_1q_2)/r^2 \hatr* dl$
where $d l \equiv d r ˆ r + r d θ ˆ θ + r sin θ d θ ˆ ϕ$ $dl \equiv dr\hatr + rd\theta \hat\theta + r sin\theta d\theta \hat\phi$

Let's make our path start and end at the same point $a$ $a$ to create a closed path:

$= \int_{a}^{a} k \frac{q_{1} q_{2}}{r^{2}} ˆ r \cdot (d r ˆ r + r d θ ˆ θ + r sin θ d θ ˆ ϕ)$ $= \int_a^a k(q_1q_2)/r^2 \hatr * (dr\hatr + rd\theta \hat\theta + r sin\theta d\theta \hat\phi)$

$= \int_{a}^{a} k \frac{q_{1} q_{2}}{r^{2}} d r$ $= \int_a^a k(q_1q_2)/r^2 dr$

$= k q_{1} q_{2} \int_{a}^{a} \frac{1}{r^{2}} d r$ $= kq_1q_2\int_a^a 1/r^2 dr$

$= - 2 k q_{1} q_{2} {[\frac{1}{r^{3}}]}_{a}^{a}$ $= -2kq_1q_2 [1/r^3]_(a)^(a)$

$= - 2 k q_{1} q_{2} (\frac{1}{a^{3}} - \frac{1}{a^{3}})$ $= -2kq_1q_2 (1/a^3 - 1/a^3)$

$= - 2 k q_{1} q_{2} (0) = 0$ $= -2kq_1q_2 (0) = 0$

We have shown that the work done by the electric force along a path starting and ending at the same point is 0. Hence, the electric force is conservative.

=======================EDIT=======================
There are three equivalent statements for showing a force is conservative:

(1) The cross product of the force vector is 0:

$\nabla \times \to F = \to 0$ $\grad xx \vec F = \vec 0$

(2) The work done by the force vector along a path that starts and ends at the same point is 0:

$W = \oint_{l} \to F \cdot d \to l = 0$ $W = \oint_l \vec F*d\vecl = 0$

(3) A force vector is the negative gradient of the potential:

$\vec F = -\grad \Phi$

I demonstrated (2) in the answer given.

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Why is an electric force conservative?

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