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quester [9]
3 years ago
14

Two identical metal balls a and b are mounted on insulating rods. Ball a has a charge of +q/2 and ball b is initially uncharged.

Ten ball a touched ball b what is resulting charge of ball a
Physics
2 answers:
oksano4ka [1.4K]3 years ago
6 0

Answer:

Help me please?

Explanation:

Did you get the answer? I believe it’s either C. +q or D. 0

rodikova [14]3 years ago
5 0

Answer:

q_end = q / 4  

Explanation:

Since the balls are in insulating barrels, it has no charge exchange with the earth, so it maintains its initial charge.

 

When the ball a touches the ball b and since the two are metallic the charge is distributed over the entire surface, by separating the two spheres that are of the same radius, the load was distributed between the two,

           

              q_end = q₀ / 2

              q_end = (q / 2) / 2

              q_end = q / 4

This is the charge on each ball.

If we keep touching the balls nothing else happens because the charges are already equal in both.

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Dmitriy789 [7]

Answer:

\omega = 1.914\ rad/s

Explanation:

Given,

Number of turns, N = 645 N

Area, A = 20.25 m²

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Maximum Emf = 1.25 V.

Angular velocity, ω = ?

Using Induced Emf formula

\varepsilon = NAB\omega

\omega= \dfrac{\varepsilon}{NAB}

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\omega = 1.914\ rad/s

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How will the electrostatic force between two electric charges change if the first charge is doubled and the second charge is onl
Vinvika [58]

Answer:

B) \frac{2}{3}

Explanation:

The electric force between charges can be determined by;

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Where: F is the force, k is the Coulomb's constant, q_{1} is the value of the first charge, q_{2} is the value of the second charge, r is the distance between the centers of the charges.

Let the original charge be represented by q, so that;

q_{1} = 2q

q_{2} = \frac{q}{3}

So that,

F = q_{1}q_{2} x \frac{k}{r^{2} }

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  = \frac{2}{3} x \frac{kq}{r^{2} }

F = \frac{2}{3} x \frac{kq}{r^{2} }

The electric force between the given charges would change by \frac{2}{3}.

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Answer:

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