- #1
Sho Kano
- 372
- 3
Homework Statement
A circular rod has a radius of curvature R = 8.11 cm, and a uniformly distributed positive charge Q = 6.25 pC and subtends an angle theta = 2.40 rad. What is the magnitude of the electric field that Q produces at the center of curvature?
Homework Equations
E = kQ/r^2
6.25 pC --> 6.25e-12 C
The Attempt at a Solution
[itex]dE\quad =\quad \frac { kdq }{ { r }^{ 2 } } \quad \quad dq=\lambda ds\\ E\quad =\quad k\lambda \int { \frac { ds }{ { r }^{ 2 } } cos\theta \quad \quad } ds=rd\theta \\ E\quad =\quad k\lambda \int { \frac { rd\theta }{ { r }^{ 2 } } cos\theta \quad \quad } \\ E\quad =\quad \frac { k\lambda }{ r } \int _{ 0 }^{ 2.4 }{ cos\theta } d\theta \quad \quad \lambda =\frac { q }{ s } =\frac { q }{ r\theta } \\ E\quad =\quad \frac { kq }{ \theta { r }^{ 2 } } sin(2.4)\\ E\quad =\quad 2.4e-4\quad N/C[/itex]
Is this the right answer? What do you guys suggest?