Unit 6 - Problem 5 ==> Electropendulum

Given:

• $$ L = 0.75 \;\mbox{meters}  $$
• $$ m = 0.02 \;\mbox{kg}  $$
• $$ E = 10 \;\mbox{n/s} $$
• $$ q = 0.01 \;\mbox{coulombs} $$

Rationale:

• $$ T = 2 \pi \sqrt\frac{L}{a} $$
• $$ F = m * a $$
• $$ a = \frac{F}{m} $$ 
• $$ T = 2 \pi \sqrt\frac{L}{\frac{F}{m} } = 2 \pi \sqrt\frac{(m * L)}{F} $$
• $$ F_e = E * q $$
• $$ T =  2 \pi \sqrt\frac{(m * L)}{(E * q)} $$

Calculate

• $$ T =  2 \pi \sqrt\frac{(0.02 * 0.75)}{(10.0 * 0.01)} $$ 
• $$ T =  2 \pi \sqrt\frac{0.015}{ 0.1} $$
• $$ T =  2 \pi \sqrt{0.15} $$ 
• $$ T =  2 \pi * 0.387298 $$ 
• $$ T = 2.43 \;\mbox{seconds} $$