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The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge q? ( k = 1/4πε 0 = 8.99 × 10 9 N ∙ m 2/C 2)

Answers

Answer 1

Answer:

Answer:

Charge, $Q=1.56* 10^(-8)\ C$

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

$E=(kQ)/(r^2)$

Q is the charge on an object

$Q=(Er^2)/(k)$

$Q=(180000\ N/C* (0.028\ m)^2)/(9* 10^9\ Nm^2/C^2)$

$Q=1.56* 10^(-8)\ C$

So, the charge on the object is $1.56* 10^(-8)\ C$ . Hence, this is the required solution.

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Two charged particles attract each other with a force of magnitude F acting on each. If the charge of one is doubled and the distance separating the particles is also doubled, the force acting on each of the two particles has magnitude (a) F/2,
(b) F/4,
(c) F,
(d) 2F,
(e) 4F,
(f) None of the above.

Answers

Answer:

F/2

Explanation:

In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant

Hence;

F= KQ1Q2/r^2 ------(1)

Where the charge on Q1 is doubled and the distance separating the charges is also doubled;

F= K2Q1 Q2/(2r)^2

F2= 2KQ1Q2/4r^2 ----(2)

F2= F/2

Comparing (1) and (2)

The magnitude of force acting on each of the two particles is;

F= F/2

Calculate the ionization potential for C+5 ( 5 electrons removed for the C atom) and in addition compute the wavelength of the transition from n=3 to n= 2.

Answers

Answer:

Ionization potential of C⁺⁵ is 489.6 eV.

Wavelength of the transition from n=3 to n=2 is 1.83 x 10⁻⁸ m.

Explanation:

The ionization potential of hydrogen like atoms is given by the relation :

$E = (13.6Z^(2) )/(n^(2) ) eV$ .....(1)

Here E is ionization potential, Z is atomic number and n is the principal quantum number which represents the state of the atom.

In this problem, the ionization potential of Carbon atom is to determine.

So, substitute 6 for Z and 1 for n in the equation (1).

$E = (13.6*(6)^(2) )/(1^(2) )$

E = 489.6 eV

The wavelength (λ) of the photon due to the transition of electrons in Hydrogen like atom is given by the relation :

$(1)/(\lambda) =RZ^(2)[(1)/(n_(1) ^(2))-(1)/(n_(2) ^(2) )]$ ......(2)

R is Rydberg constant, n₁ and n₂ are the transition states of the atom.

Substitute 6 for Z, 2 for n₁, 3 for n₂ and 1.09 x 10⁷ m⁻¹ for R in equation (2).

$(1)/(\lambda) =1.09*10^(7) *6^(2)[(1)/(2 ^(2))-(1)/(3 ^(2) )]$

$(1)/(\lambda)$ = 5.45 x 10⁷

λ = 1.83 x 10⁻⁸ m

If you see Sagittarius high in your night sky on June 20 and today is your birthday, what is your zodiac constellation?

Answers

Answer:

Gemini

Explanation:

Zodiac Constellation or Sun sign is the constellation in which the Sun resides on the date of birth of a person. Throughout the year the Sun crosses across 12 constellations thus there are 12 Sun-signs. Though astronomically the Sun crosses across 13 constellations so there should be 13 zodiacs but most of the astrologers do not accept this. On the date of June 20, the Sagittarius which is a summer constellation and a zodiac can be seen high up in the sky in the night time. At this time the Sun will be in a constellation which is almost opposite to Sagittarius in the celestial sphere. That constellation is Gemini. Thus for a person born on 20 June, zodiac will be Gemini.

A laser (electromagnetic wave) has the maximum electric field strength of 1.0x1011 V/m. What is the force the laser applies on a mirror (totally reflective) of 5.0 mm2 area? A. 2.76 x105N B. 1.21 x106N C. 1.94 x106N D.4.43 x105 N E. 7.82 x104N