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Mashcka [7]
2 years ago
7

A positively charged sphere with a charge of 8Q is separated from a negatively charged sphere -2Q by a distance r. There is an a

ttractive force F exerted on each sphere. The spheres briefly touch each other and move to the original distance r. What is the new force on each sphere in terms of F
Physics
1 answer:
djverab [1.8K]2 years ago
4 0

Answer:

Explanation:

For the first case , the expression for electrostatic force can be given by the following .

F = K x 8Q x 2Q / r² where k is a constant .

F = K 16 Q² / r²

When they touch , some charge is neutralized . Net charge remaining

= 8Q - 2 Q = 6 Q

Charge on each sphere = 6Q/2 = 3 Q .

Force between them

F₁ = k 3Q x 3 Q / r² = k 9 Q² / r²

F₁ / F = 9 / 16

F₁ = 9 F / 16 .

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if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu
RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

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

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

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