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attashe74 [19]
3 years ago
13

A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±

Q.
Part A
What is the ratio Ef/Ei of the final electric field strength Ef to the initial electric field strength Ei if Qis doubled?
Part BWhat is the ratio Ef/Ei of the final electric field strength Ef to the initial electric field strength Ei if Lis doubled?Part CWhat is the ratio Ef/Ei of the final electric field strength Ef to the initial electric field strength Ei if d is doubled?
Physics
1 answer:
vampirchik [111]3 years ago
3 0

Answer: A) 2 B) 4 C) 1

Explanation:

The Electric field from a parallel-plate capacitor  is given by:

A) E=Q/(L^2 * ε0) so if we put a charge double the final electric field is double that the original.

B) from the above expression for the electric field,  If the size of the plate is double, then the E final is four times weaker that the original.

C) If the distante between plates is doubled the final electric field is the same that initial.

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Explanation:

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How do I find the x and y components and the resultant force?​
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Explanation:

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3 years ago
You hold a ruler that has a charge on its tip 6 cm above a small piece of tissue paper to see if it can be picked up. The ruler
Fed [463]

Answer:

9.81 × 10⁻¹⁰ C

Explanation:

Given:

Distance between the tissue and the tip of the scale, r = 6 cm = 0.06 m

Charge on the ruler, Q = - 12 μC = - 12 × 10⁻⁶ C

Mass of the tissue = 3 g = 0.003 Kg

Now,

The force required to pick the tissue, F = mg

where, g is the acceleration due to gravity

also,

The force between (F) the charges is given as:

F=\frac{kQq}{r^2}

where,

q is the charge on the tissue

k is the Coulomb's constant = 9 × 10⁹ Nm²/C²

thus,

mg=\frac{kQq}{r^2}

on substituting the respective values, we get

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or

q = 9.81 × 10⁻¹⁰ C

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