The potential as a function of position in a region is \(V(x)=\) \(3 x-2 x^{2}-x^{3},\) with \(x\) in meters and \(V\) in volts. Find (a) all points on the \(x\) -axis where \(V=0,\) (b) an expression for the electric field, and (c) all points on the \(x\) -axis where \(E=0\)

Short Answer

Expert verified
The potential, V, is zero at \(x=0\) (only real root). The electric field, E, is zero at \(x=0\) and \(x=-2/3\)

Step by step solution

01

Find the roots of potential function

The equation given is \(V(x)=3x-2x^2-x^3\). We first set this equal to zero to find out when the potential is 0. This gives the equation \(0=3x-2x^2-x^3\), or equivalently \(x^3 - 2x^2 + 3x = 0\). The roots of this cubic equation represent the x-values where the potential is zero.
02

Factor the cubic equation

The equation \(x^3 - 2x^2 + 3x = 0\) can be factored by taking an x out of the expression, resulting in \(x*(x^2 - 2x +3) = 0\). Therefore the potential becomes zero when x=0, or when \(x^2 - 2x + 3 = 0\). However, the quadratic equation \(x^2 - 2x +3 =0\) does not have any real roots, so the only real solution for \(V=0\) is at \(x=0\)
03

Find the electric field

The electric field E is the negative derivative of the potential function. This means we differentiate \(V(x)=3x-2x^2-x^3\), with respect to x and then change the sign. This gives us \(E(x)=-\frac{dV}{dx}=2x+3x^2\)
04

Find when electric field is zero

Setting \(E(x)=0\), we get \(2x+3x^2=0\). Factoring out an \(x\) gives \(x*(2+3x)=0\), which implies that E(x)=0, at x=0 and x=-2/3, these points are the solutions to the questions.

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