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ArbitrLikvidat [17]
2 years ago
10

Two identical conducting spheres are charged with a net charge of +5.0 q on the first sphere and a net charge of −8.0 q on the s

econd sphere. The spheres are brought together, allowed to touch, and then separated. What is the net charge on each sphere now?
Physics
1 answer:
monitta2 years ago
7 0

Answer:

The net charge on each sphere is -1.5 q

Explanation:

Conductors are materials that allow the electrons which are the carriers of the charges to move between them, and when two conductors come in contact, the available charge is shared by the two conductors and the resultant like charges will spread on the surface of the conductor due to the repellent effect between similar charges such that if the conductors are identical, the resultant charge becomes evenly shared by the conductors when they become separated again

The given parameters of the conducting spheres meant to touch are;

The net charge on the first sphere, Q₁ = +5.0 q

The net charge on the second sphere, Q₂ = -8.0 q

The net charge on each sphere after touching and then separated, 'Q', is given as follows;

Q = \dfrac{Q_1 + Q_2}{2}

Therefore, by substituting the known values of the variables, we have;

Q = \dfrac{5 \ q+ (-8 \ q)}{2} = -\dfrac{3 \ q}{2} = -1.5 \ q

The net charge on each sphere, Q = -1.5 q.

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Umnica [9.8K]

Explanation:

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3 years ago
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The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able
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Answer:

-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

q_e=-1.6\cdot 10^{-19}\ C

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is 5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C

Let's test the possible charges listed in the question:

-8.0 \times 10 ^{-19 }. We have just found it's a possible charge of a particle

-3.2 \times 10 ^{-19 }. Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

-1.2 \times 10 ^{-19 } this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 } cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge -4.8 \times 10 ^{-19 }\ C is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C

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3 years ago
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

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