Final answer:
The length of time required to dissipate 87% of the initially stored energy is approximately 0.029 seconds.
Step-by-step explanation:
To find the length of time required to dissipate 87% of the initially stored energy, we can use the formula for exponential decay. The energy stored in the capacitor is given by E(t) = 0.5Cv(t)^2, where C is the capacitance and v(t) is the voltage at time t. We are given v(t) = 200e^(-t/0.02) and we want to find the time t when the energy is 87% of its initial value. Let's denote the initial energy as E(0) = 0.5Cv(0)^2 and the energy at time t as E(t) = 0.5Cv(t)^2.
Since we want to find the time when the energy is 87% of its initial value, we can write the equation:
0.5Cv(t)^2 = 0.87 * 0.5Cv(0)^2
Dividing both sides by 0.5C, we get:
v(t)^2 = 0.87 * v(0)^2
Taking the square root of both sides gives:
v(t) = sqrt(0.87) * v(0)
Now we can substitute the given values: v(t=0) = 200V, v(t) = sqrt(0.87) * 200V. Solving for t, we have:
t = -0.02 * ln(200 / (sqrt(0.87) * 200))
t = -0.02 * ln(1 / sqrt(0.87))
t ≈ 0.029 s