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電磁気の基礎 課題 001

Last updated at Posted at 2024-11-06

問題文

1 Cの電気量をもち、1 m離れている2つの電荷の間に働く電気力は、約何 kgの質量の物体に働く重力に等しいか。


解説

  1. 電気力の計算
    2つの電荷の間に働く電気力 ( F ) はクーロンの法則を使って求められます。

$$ F = \frac{k \cdot q_1 \cdot q_2}{r^2} $$

ここで、
$ ( k = 8.99 \times 10^9 , \text{N} \cdot \text{m}^2/\text{C}^2 )$(真空中のクーロン定数)
$( q_1 = q_2 = 1 \, \text{C} )$(各電荷の電気量)
$( r = 1 , \text{m} )$(距離)

代入すると、
$$
F = \frac{8.99 \times 10^9 \times 1 \times 1}{1^2} = 8.99 \times 10^9 \, \text{N}
$$

  1. 重力との比較
    次に、質量 ( m ) の物体に働く重力 ( F_g ) を次の式で求めます。
    $$
    F_g = m \cdot g
    $$
    ここで、$( g = 9.8 , \text{m/s}^2 )$(重力加速度)とし、$( F_g = F )$ とすると、
    $$
    m = \frac{F}{g} = \frac{8.99 \times 10^9}{9.8} \approx 9.17 \times 10^8 , \text{kg}
    $$

答え

この電気力は、約 $( 9.17 \times 10^8 \, \text{kg} )$(約9億1700万 kg)の質量に働く重力に等しいです。

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