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Old Sep 29th 2019, 09:21 AM   #1
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Post Question about relationship between Acceleration and Electric Potential

Hi everyone. Im a Engineering student and i having doubts about this question:

"What is the acceleration vector $\displaystyle a(x,y,z)$ of a particle of mass $\displaystyle m0$ and charge $\displaystyle q0$ when there is a Potential vector$\displaystyle V(x,y,z) = c0 x3 c1 + x-5 y5 z4$ where '$\displaystyle c0$' and '$\displaystyle c1$' are constants."

Well I started with the relation

$\displaystyle V = -W/q0 $ (1)

and $\displaystyle W = F d$ (being F and d vectors) (2)

also $\displaystyle F = m0*a$ (being 'a' a vector) (3)

So I ended up with this equation that relates a and V replacing equation (3) in (2):

$\displaystyle W = (m0*a) d $(4)

And replacing (4) in (1) I have:

$\displaystyle V = - m0*a*d / q0 $(5)

Which means:

$\displaystyle a = - V*q0 / m0*d$ (6)

Here I got stuck and don't know how to proceed. I appreciate very much any help.
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Old Sep 30th 2019, 05:59 AM   #2
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There's no need to introduce work done here. The potential is related to the electric field strength by:

$\displaystyle \vec{E} = - \nabla V$

Assuming the scalar field $\displaystyle V(x,y,z) = c_0 c_1 x^3 + x - 25 yz^4$, we can compute the gradient, $\displaystyle \nabla V$.

$\displaystyle \nabla V = \frac{\partial V}{\partial x} \hat{x} + \frac{\partial V}{\partial y} \hat{y} + \frac{\partial V}{\partial z} \hat{z}$
$\displaystyle = \left(3 c_0 c_1 x^2 +1\right) \hat{x} - 25z^4 \hat{y} - 100 y z^3 \hat{z}$

Assuming that the only force in the problem is the electric force, we have

$\displaystyle \vec{F} = \vec{E} q_0 = m_0 \vec{a}$.

Therefore,

$\displaystyle \vec{a} = \frac{\vec{E} q_0}{m_0} = \frac{-q_0 \nabla V }{m_0}$
$\displaystyle = \frac{q_0}{m_0} \left(-\left(3 c_0 c_1 x^2 +1\right) \hat{x} + 25z^4 \hat{y} + 100 y z^3 \hat{z}\right)$

You might want to double-check that the potential you wrote down is the same as the one I assumed...
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