Answer :
Answer:
a) [tex]\bf f(x,y,z)=200ln(x^2+z^2)[/tex]
b) [tex]\bf 3.9819*10^{16}\;m/seg[/tex]
Step-by-step explanation:
The electric field can be written as
[tex]\bf F(x,y,z)=(\displaystyle\frac{400x}{x^2+z^2},0,\displaystyle\frac{400z}{x^2+z^2})[/tex]
a)Find a potential function for F
A potential function for F would be a function f such that
[tex]\bf \nabla f = F[/tex]
That is, the gradient of f equals F.
If such a function exists then
[tex]\bf \displaystyle\frac{\partial f}{\partial x}=\displaystyle\frac{400x}{x^2+z^2}\\\\\displaystyle\frac{\partial f}{\partial y}=0\\\\\displaystyle\frac{\partial f}{\partial z}=\displaystyle\frac{400z}{x^2+z^2}[/tex]
Integrating the first equation with respect to x
[tex]\bf f(x,y,z)=\displaystyle\int\displaystyle\frac{\partial f}{\partial x}dx+h(y,z)[/tex]
where h(y,z) is a function that does not depend on x. We have then
[tex]\bf f(x,y,z)=\displaystyle\int\displaystyle\frac{400x}{x^2+z^2}dx+h(y,z)=400\displaystyle\int\displaystyle\frac{x}{x^2+z^2}dx+h(y,z)=\\\\200ln(x^2+z^2)+h(y,z)[/tex]
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
[tex]\bf \boxed{f(x,y,z)=200ln(x^2+z^2)}[/tex]
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)
[tex]\bf W=200ln(1^2+1^2)-200ln(5^2+7^2)=-722.183583[/tex] 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
[tex]\bf W=-\displaystyle\frac{m(v_Q)^2}{2}[/tex]
where
m = mass of an electron
[tex]\bf v_Q[/tex] = speed at point Q
Replacing in the equation
[tex]\bf -722.183583=-\displaystyle\frac{9.10938356*10^{-31}(v_Q)^2}{2}\Rightarrow\\\\\Rightarrow v_Q=\sqrt{\displaystyle\frac{2*722.183583}{9.10938356*10^{-31}}}=3.9819*10^{16}\;m/seg[/tex]