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3 years ago

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Lightning As a crude model for lightning, consider the ground to be one plate of a parallel-plate capacitor and a cloud at an al

titude of 650 m to be the other plate. Assume the surface area of the cloud to be the same as the area of a square that is 0.70 km on a side. a. What is the capacitance of this capacitor?b. How much charge can the cloud hold before the dielectric strength of the air is exceeded and a spark (lightning) results?

1 answer:

olga nikolaevna [1]3 years ago

7 0

Answer:

$6.67154\times 10^{-9}\ F$

13.009503 C

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

k = Dielectric constant of air $3\times 10^6\ V/m$

Side of plate = 0.7 km

A = Area

d = Distance = 650 m

Capacitance is given by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 700^2}{650}\\\Rightarrow C=6.67154\times 10^{-9}\ F$

The capacitance is $6.67154\times 10^{-9}\ F$

Electric field is given by

$Q=CV\\\Rightarrow Q=Ckd\\\Rightarrow Q=6.67154\times 10^{-9}\times 3\times 10^6\times 650\\\Rightarrow Q=13.009503\ C$

The charge on the cloud is 13.009503 C

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If $g = 9.81\; \rm m \cdot s^{-2}$ , the potential energy of the block of $m = 2.20\; \rm kg$ would be $m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J$ when it was at the top of the incline.

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$\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}$ .

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

$p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

$p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right before the collision: approximately $18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right after the collision: $(m\cdot v_m + m \cdot v_M)$ .

For momentum to conserve in this collision, $v_m$ and $v_M$ should ensure that $m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Kinetic energy of the two blocks right before the collision: approximately $77.695\; \rm J$ and $0\; \rm J$ . Sum of these two values: approximately $77.695\; \rm J\!$ .

Sum of the energy of each block right after the collision:

$\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)$ .

Similarly, for kinetic energy to conserve in this collision, $v_m$ and $v_M$ should ensure that $\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J$ .

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$\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.$ .

Solve for $v_m$ and $v_M$ (ignore the root where $v_M = 0$ .)

$\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.$ .

The collision flipped the sign of the velocity of the $m = 2.20\; \rm kg$ block. In other words, this block is moving backwards towards the incline after the collision.

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