Question

An electron at rest at P = (5,3,7) moves along a path ending at Q = (1, 1, 1) under the influence of the electric field (in newtons per coulomb)

F(x, y, z) = 400(x^2 + z^2)^-1 (x, 0, z).

a)Find a potential function for F

b) What is the electron’s speed at the point Q? (Use conservation of energy and the value qe/me = −1.76 × 1011 C/kg, where qe and me are the charge and mass on the electron, respectively

Answer 1

a)
`\bf f(x,y,z)=200ln(x^2+z^2)`

b)
`\bf 3.9819*10^(16)\;m/seg`

Explanation:

The electric field can be written as

`\bf F(x,y,z)=(\displaystyle(400x)/(x^2+z^2),0,\displaystyle(400z)/(x^2+z^2))`

a)Find a potential function for F

A potential function for F would be a function f such that

`\bf \\abla f = F`

That is, the gradient of f equals F.

If such a function exists then

`\bf \displaystyle(\partial f)/(\partial x)=\displaystyle(400x)/(x^2+z^2)\\\\\displaystyle(\partial f)/(\partial y)=0\\\\\displaystyle(\partial f)/(\partial z)=\displaystyle(400z)/(x^2+z^2)`

Integrating the first equation with respect to x

`\bf f(x,y,z)=\displaystyle\int\displaystyle(\partial f)/(\partial x)dx+h(y,z)`

where h(y,z) is a function that does not depend on x. We have then

`\bf f(x,y,z)=\displaystyle\int\displaystyle(400x)/(x^2+z^2)dx+h(y,z)=400\displaystyle\int\displaystyle(x)/(x^2+z^2)dx+h(y,z)=\\\\200ln(x^2+z^2)+h(y,z)`

By taking the partial derivatives with respect y and z we can notice that h(z,y)=0.

Therefore, a potential function for F is

`\bf \boxed{f(x,y,z)=200ln(x^2+z^2)}`

b) What is the electron’s speed at the point Q?

Since F is conservative, the work done to move the particle from P to Q does not depend on the path, but only on the potential function f.

Let W be the work done when moving the particle from P to Q

W = f(Q) - f(P) = f(1,1,1) - f(5,3,7)

`\bf W=200ln(1^2+1^2)-200ln(5^2+7^2)=-722.183583` joules

According to the law of conservation of energy, the work done to move the electron from P to Q equals the change in kinetic energy of the object.

Since the electron is at rest in P, the kinetic energy at P equals 0 and we have

`\bf W=-\displaystyle(m(v_Q)^2)/(2)`

where

m = mass of an electron

`\bf v_Q` = speed at point Q

Replacing in the equation

`\bf  -722.183583=-\displaystyle(9.10938356*10^(-31)(v_Q)^2)/(2)\Rightarrow\\\\\Rightarrow v_Q=\sqrt{\displaystyle(2*722.183583)/(9.10938356*10^(-31))}=3.9819*10^(16)\;m/seg`