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3 years ago

All of the orbitals in a given subshell have the same value of the __________ quantum number.

1 answer:

3 0

All of the orbitals in a given subshell have the same value of the "<span>magnetic and principal" quantum number

Hope this helps!</span>

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3 years ago

Find the distance separating two dust particles (in m) if each has a charge of +e and the Coulomb force between them has magnitu

marta [7]

Answer:

Distance between dust particles, $r=1.2\times 10^{-7}\ m$

Explanation:

Given that,

Charge on the dust particle, $q_1=q_2=1.6\times 10^{-19}\ C$

Force between dust particles, $F=1.6\times 10^{-14}\ N$

We need to find the distance between two dust particles. The electrostatic force between dust particles is given by :

$F=k\dfrac{q_1q_2}{r^2}$ , r = the distance

$r=\sqrt{k\dfrac{q_1q_2}{F}}$

$r=\sqrt{9\times 10^9\times \dfrac{(1.6\times 10^{-19})^2}{1.6\times 10^{-14}}}$

$r=1.2\times 10^{-7}\ m$

So, the distance between two dust particles is $1.2\times 10^{-7}\ m$ . Hence, this is the required solution.

7 0

2 years ago

35. A spherical conductor has a radius of 14.0 cm and a charge of 26.0 mC. Calculate the electric field and the electric poten-

bulgar [2K]

Answer:

Therefore,

$V_{10}=2.34\times 10^{6}\ V$

$V_{20}=1.17\times 10^{6}\ V$

$V_{14}=1.67\times 10^{6}\ V$

Explanation:

Given:

A spherical conductor has a radius of 14.0 cm

Q = 26.0 μC ( Assume it to be microC in the question it is miliC )

To Find:

Electric potential at

(a) r 5 10.0 cm, (b) r 5 20.0 cm, and (c) r 5 14.0 cm from the center.

Solution:

Electric potential due to point charge Q at any distance r from the charge is,

$V=\dfrac{kQ}{r}$

Where,

V = Electric Potential

k = Coulombs constant = 9 × 10⁹ Nm²/C².

Q = Charge

r = distance in meter

Substituting the values we get

At r = 10 cm = 0.1 m

$V=\dfrac{9\times 10^{9}\times 26\times 10^{-6}}{0.1}$

$V=2.34\times 10^{6}\ V$

At r = 20 cm = 0.2 m

$V=\dfrac{9\times 10^{9}\times 26\times 10^{-6}}{0.2}$

$V=1.17\times 10^{6}\ V$

At r = 14 cm = 0.14 m

$V=\dfrac{9\times 10^{9}\times 26\times 10^{-6}}{0.14}$

$V=1.67\times 10^{6}\ V$

Therefore,

$V_{10}=2.34\times 10^{6}\ V$

$V_{20}=1.17\times 10^{6}\ V$

$V_{14}=1.67\times 10^{6}\ V$

7 0

3 years ago