Question

Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charge of the charge transferred I s30 C? (a )What is the charge in energy of that transferred charge? (b) If all the energy released could be used to accelerate a 1000kg car from rest. What would be its final speed?

Answer 1

Answer:

a) $U_{e} = 3 \times 10^{10}\,J$, b) $v \approx 7745.967\,\frac{m}{s}$

Explanation:

a) The potential energy is:

$U_{e} = Q \cdot \Delta V$

$U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)$

$U_{e} = 3 \times 10^{10}\,J$

b) Maximum final speed:

$U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }$

The final speed is:

$v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }$

$v \approx 7745.967\,\frac{m}{s}$

Answer 2

The change in the energy transferred or potential energy is 3 x 10¹⁰ J.

The final speed of the charge released is 7745.97 m/s.

Change in energy transferred

The change in the energy transferred or potential energy is calculated as follows;

U = QV

U = 30 x 1 x 10⁹

U = 3 x 10¹⁰ J

Speed of the charge released

The speed of the charge released is calculated by applying law of conservation of energy.

$K.E = U\\\\\frac{1}{2} mv^2 = U\\\\v= \sqrt{\frac{2U}{m} } \\\\v = \sqrt{\frac{2\times 3\times 10^{10} }{1000} }\\\\v = 7745.97 \ m/s$

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