Calculating Work Required to Move a Point Charge of +8μC from (0,4m) to (3m,0)

[jan2905]

A point charge, Q=+1mC, is fixed at the origin. How much work is required to move a charge, q=+8(micro)C, from the point (0,4meters) to the point (3meters,0)?



I was unable to derive anything.



I guessed at around 40J. Correct? If so HOW! lol

 

[Feldoh]

How are electric potential and work related?

 

[jan2905]

Potential=E*d

 

[Feldoh]

You have there an equation relating the electric potential to the electric field and the separation between the change in potential. While it's useful you need to find a way to relate V to W...

 

[jan2905]

I don't know... V=Eq*d ? because F=Eq ?

 

[Feldoh]

Ok so it V = E * d then it seems logical to say that

V = (F*d)/q, right? Now the nice thing about electric potential is that it is a scalar.

 

[chrisk]

Since the electic field created by Q is not uniform E(x) must be used

E(x)=K\frac{Q}{x^2}

where K is

\frac{1}{4\pi\varepsilon_0}

The force on q is

F(x)=qE(x)

So, the work done on q is

W=-\int\mbox{F(x)dx}

 

[jan2905]

This problem doesn't require integration.

and... how did you derive V = (F*d)/q? if F=Eq, then by your derivation V=(Eq)*d/q, which is reduced to V=Ed? how does V=Eqd=Ed?

 

[Feldoh]

Typing error... V=E*d => V = F*d / q

 

[chrisk]

Integration is a way to solve the problem. Simply choose a convenient path starting at (0,4) and ending at (3,0). Any path will work because E is a conservative field. Using potential is another way. Finding the potential difference between the two points then multiplying this difference by q will give the desired result. Because E is not uniform use integration.

V=-\int_{\infty}^{r}\mbox{E(r)dr}=\frac{Q}{4\pi \varepsilon_0\mbox{r}}

V_1=\int_{4}^{\infty}\mbox{E(y)dy}=\frac{Q}{4\pi \varepsilon_0\mbox{4}}

V_2=\int_{3}^{\infty}\mbox{E(x)dx}}=\frac{Q}{4\pi \varepsilon_0\mbox{3}}

W=(V_2-V_1)q