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Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 12.0 mg piece of tape held 1.00 cm above another.

Sagot :

oladayoademosu

Answer:

36.1 nC

Explanation:

The electrostatic force F = kqq'/r² since q = q', F = kq²/r² where q = charge on tape, r = distance between tapes = 1.00 cm = 1 × 10⁻² m and k = 9 × 10⁹ Nm²/C².

Given that F = weight of 12.0 mg piece of tape, F = mg where m = mass of tape = 12.0 mg = 12 × 10⁻³ kg and g = acceleration due to gravity = 9.8 m/s²

So, kq²/r² = mg

q²/r² = mg/k

q² = mgr²/k

taking square-root of both sides,

q = √(mg/k)r

So, substituting the values of the variables into the equation, we have

q = √(mg/k)r

q = √(12 × 10⁻³ kg × 9.8 m/s²/ 9 × 10⁹ Nm²/C²)1 × 10⁻² m

q = √(117.6 × 10⁻³ kgm/s²/9 × 10⁹ Nm²/C²)1 × 10⁻² m

q = √(13.07 × 10⁻¹² C²/m²)1 × 10⁻² m

q = 3.61 × 10⁻⁶ C/m × 1 × 10⁻² m

q =  3.61 × 10⁻⁸ C

q = 36.1 × 10⁻⁹ C

q = 36.1 nC

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