Voltage = current x resistance
since R is doubled, current must reduce by half.
So,
new current = 120/2 = 60mA
We know, F = 1/4πε * q₁q₂ / r²
Here, q₁ = 5 * 10⁻⁶ C
q₂ = 2 * 10⁻⁶ C
r = 3 * 10⁻² m west
Substitute their values,
F = (9 * 10⁹) (5 * 10⁻⁶) (2 * 10⁻⁶) / (3 * 10⁻²)²
F = 100 N [ East of positive charge ]
Hope this helps!
The total capacitance is <em>C</em> such that
1/<em>C</em> = 1/(5.0 µF) + 1/(14 µF) + 1/(21 µF)
Solve for <em>C</em> :
<em>C</em> = 1 / (1/(5.0 µF) + 1/(14 µF) + 1/(21 µF)) ≈ 3.1 µF
Answer:
carbon dioxide i think please be right
