# Question about relationship between Acceleration and Electric Potential

#### rayonnunes

"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.

#### benit13

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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