Recent content by Roverse

Yeah I'm still learning latex, misplaced a curly brace. One moment. Is it good now?

Quick correction: I didn't correctly write the solution integral on here even though I did in my own calculations...

Hi I'm Robbie, a uni student studying physics and mathematics. I hate experiments and I like theory. Also, I'm not passionate about physics, yet. I really enjoy investigating math maybe because it has a much lower barrier for discovery (and breadth) in comparison to physics. Anyways, I hope I...

I understand the way the reference did it. I actually just found the error (I forgot to divide by a factor of 2 (for my little half distance across the height, where before I had the entire). My answer is correct now. Thanks for the help!

Ah yeah that was a typo (out of confusion, I thought you wanted the value if a point charge was at the bisector). In my above equation. the r is actually $$\sqrt{\left(\left|\frac{d}{2}-x\right|\right)^2+\left(d-\left(\frac{d}{2}\right)^2\right)}$$. where x goes from 0 to 18cm along the rod. I...

Because $$\vec{E}_{perp}=\frac{kq}{r^2}$$, and at the perpendicular bisector $$r=\sqrt{\left(\frac{d} {2}\right)^2+d^2} = \frac{9\sqrt{3}}{2}$$, the expresion is then $$dqdx\cdot\frac{k}{\frac{9\sqrt{3}}{2}^2}$$. Though I am not sure I fully understand what you mean, since I'm unsure why it...

Homework Statement Three 18-cm long rods form an equilateral triangle. Two of the rods are charged to +10 nC, and the third to - 10 nC. What is the electric field strength at the center of the triangle? Homework Equations $$ \vec{E} = \frac{k*q}{r^2} $$ The Attempt at a Solution 1. Draw...