- #1
CraigH
- 222
- 1
Homework Statement
I'm pretty sure I have done questions 1 to 4 correctly, it is just question 5 I am struggling with.
I have an exam in one week and this is revision so it will not get marked by a tutor, and it would really help if someone could give my answers a quick check and help me with question 5.
Thanks!
1)
a) Calculate the electric field strength at position (2,3,0) of a 1 coulomb point charge placed at position (0,0,0)
2)
a) Calculate the electrical potential at point (2,3,0)
b) Calculate the electrical potential at point (0,0,0)
c) Calculate the electrical potential difference
e) Calculate the electrical potential difference between (x,y,z) and (x1,y1,z1)
3)
a) A 0.3 Coulomb point charge is placed at (2,3,0). Calculate the force on this charge
4)
a) Calculate the electrical potential energy of this point charge.
5)
a) If the charges mentioned were not point charges, how would the equations and answers change.
b) Derive the equations needed for these questions if the charges where not point charges, but instead:
i) Infinitely long cylindrical parallel wires.
ii) Infinitely long X mm thick parallel micro strip wires
ii) An infinitely long co-axle cable with the inner conductor at charge q and the outer at charge Q
iii) Two infinitely large parallel plates
c) How would these equations change if the objects where of a finite length.
Homework Equations
1)
[itex]\underline{F} = \frac{qQ}{\underline{r}^2} * \frac{1}{4\pi\epsilon_{0}}[/itex] * [itex]\underline{\widehat{r}}_{qQ}[/itex]
[itex]\underline{\widehat{r}}_{qQ}[/itex] Is the unit vector in the direction to get from q to Q
2)
[itex]\underline{E} = \frac{Q}{\underline{r}^2} * \frac{1}{4\pi\epsilon_{0}}[/itex] * [itex]\underline{\widehat{r}}[/itex]
3)
[itex]\underline{F} = \underline{E}q[/itex]
4)
[itex]\underline{V} = - \oint\underline{E}.dl [/itex]
The Attempt at a Solution
1a)
r=SQRT(2^2+3^2) = 3.6
direction of r is pointing perpendicularly away from the point charge.
r unit vector = (2i+3j)/3.6
E=1/(3.6^2) * 8.99*10^9 * (2i+3j)/3.6
E = 6.94*10^8 * (2i+3j)/3.6
E =(3.85*10^8)i + (5.78*10^8)j
2a)
V = E * 3.6
V = (1.39*10^9)i + (2.08*10^9)j
2b)
V = 0
2c)
V = (1.39*10^9)i + (2.08*10^9)j
2e)
will be done exactly the same but with letters instead of numbers.
3a)
F = (3.85*10^8)i + (5.78*10^8)j * 0.3
F= (1.155*10^8)i + (1.734*10^8)j
4a)
Energy = (1.155*10^8)i + (1.734*10^8)j * 3.6
Energy = (4.158*10^8)i + (6.2424*10^8)j
Energy = SQRT(a^2+b^2)
Energy = 7.50*10^8 J
(should I have integrated the force across the distance instead of just multiplying them?)
5a) I have absolutely no clue here. I think I need to derive the equations I gave in the beginning from a more general form. Probably from maxwell's equations. can someone please help me out on this bit?
Thank you!