If I have a space of uniform electric field, let's say always pointing upwards and of strength 5 N/C, if I place a ball of 2 C in that field. It should experience a force of 10 N and accelerate upwards, right? (The space is also free from gravitational field)

I have a few questions here,

  1. Would the field of the 2 C ball mess up the field of the space? Would it cause problems (as in, will the ball actually experience a lesser force because of this or something?)

  2. I heard that accerlerating charges give out radiation and loose energy. What will happen here, the ball won't keep accelerating away as I initially thought?

  3. Moreever, that statement "accelerating charges give out radiation and loose Energy" is confusing me. Does this mean it's all frame dependent? If I rub my hair with a balloon, it will get charged..If I after rubbing the balloon keep it aside and start running..The charges should loose energy and emit radiation? This is really confusing me..Is it really frame dependent like that?

  • $\begingroup$ When a charge accelerates there is always a self force called the Abraham–Lorentz force which is experienced by the particle. It is exactly this force which lowers the kinetic energy of the particle, resulting in it losing energy due to radiation. Remember that in order for the particle to lose its kinetic energy there has to be a force that makes it decelerate because of Newton's second law. You may want to check out this link: en.wikipedia.org/wiki/Abraham%E2%80%93Lorentz_force $\endgroup$ – Sahand Tabatabaei Nov 24 '17 at 21:31
  • $\begingroup$ As for question 3, remember that energy itself is a frame-dependent quantity. $\endgroup$ – probably_someone Nov 24 '17 at 21:54

Although the question does not say so explicitly, the assumption is that the $\vec E$-field is constant regardless of the additional charge you place in it. One way to generate this type of constant field is to use a capacitor where the distance between the plates is much less than the dimensions of the plate themselves: the field near the centre of the capacitor is approximately constant in magnitude and direction. Since an "external agent" presumably maintains the voltage between the plates at some constant value, inserting a charge in between the plates would not affect the force on the charge.

Most of the elementary electrostatics is in fact "quasi-statics", meaning you can neglect many effects like radiation loss when the velocities are small (compare to speed of light) or accelerations are small.

For small velocities $(v/c\ll 1)$, radiation is predominantly perpendicular to the acceleration and the total radiation power is approximately $$ P\approx \frac{1}{4\pi\epsilon_0}\frac{2q^2 \dot{\beta}^2}{3c} \tag{1} $$ where $$ \dot\beta=\frac{d}{dt}\frac{v}{c}=\frac{a}{c}\, . $$ Eq.(1) is the Larmor formula for radiation due to a non-relativistic charge. Eq.(1) goes like $\dot{\beta}^2$ so unless the acceleration is large (in which case you could rapidly reach velocities for which the approximation $v/c\ll 1$ no longer holds), the power will remain generally small.

[For arbitrary $v/c$, the full radiation power is obtained by multiplying (1) by $\gamma^6$,where $\gamma=1/\sqrt{1-v^2/c^2}$.]

  • $\begingroup$ My question was that the uniform electric field was already present. (Say from an infinite charged surface below), if a new ball with significant charge is introduced, will F = qE still remain valid (or not because of the field from the ball distorting the uniform field previously there?) $\endgroup$ – Rick Nov 25 '17 at 6:25
  • 1
    $\begingroup$ @Rick yes of course. $\vec F=q\vec E$ is the force due to the external electric field on a point charge, irrespective of $q$. $\endgroup$ – ZeroTheHero Nov 25 '17 at 6:32

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