Yaadannoo boqonnaa fi gaaffii shaakalaa

Boqonnaa 6 Fiizikii kutaa 11: Electrostatics fi Electric Circuit

Wanti fuula kana irra jiru hundi boqonnaa 6 kitaaba Fiizikii kutaa 11 tiif: wanta boqonnaan haguugu, yaadannoo gabaabaa, gaaffii shaakalaa deebii isaanii waliin, fi viidiyoo barnootaa bakka jirutti.

3
Kaardii: 10
Gaaffii shaakalaa boqonnaa 6: 217–286
Fuula kitaabaa boqonnaa 6

Yaadannoo gabaabaa boqonnaa 6

Kaardiiwwan fuula Fiizikii kutaa 11 irratti boqonnaa kana jala galmaa'an. Tokko tokkoon bani.

Seera CoulombFoormulaa

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

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

$F$: guddina humna elektirostaatikii chaarjii qabxii lamaan gidduu jiru $N$
$k$: Yeroo Coulomb, k = 1/(4 pi epsilon_0) = 9.0 x 10^9 $N\,m^2\,C^{-2}$
$q_1, q_2$: chaarjii qabxii lamaan $C$
$r$: fageenya chaarjii lamaan gidduu jiru $m$

Yoom akka itti fayyadamtu

Yeroo humna elektirostaatikii chaarjii qabxii lamaan fageenya r irraa adda ta'an gidduu barbaaddan kana fayyadamaa.

Of eeggadhu

Fageenya karra'uu dura gara meetiraatti jijjiiruu dhiyeessaa. Fakkeenyaaf 1.0 cm = 0.01 m dha, kanaaf r^2 = 1.0 x 10^-4 m^2 dha, 1.0 miti. Walitti harkisuu ykn walitti dhiibuun kan murtaa'u mallattoo chaarjii irraa, ulaagaa irraa miti.

Grade 11 Physics, fuula 217-286 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Dirree elektirikii chaarjii qabxiiFoormulaa

E = k\frac{Q}{r^2}

E = k\frac{Q}{r^2}

$E$: cimina dirree elektirikii fageenya r irraa chaarjii qabxii $N\,C^{-1}$
$k$: Yeroo Coulomb $N\,m^2\,C^{-2}$
$Q$: chaarjii madda ta'e dirree uumu $C$
$r$: fageenya chaarjii maddaa irraa hanga qabxii barbaadamaa $m$

Yoom akka itti fayyadamtu

Yeroo cimina dirree elektirikii qabxii fageenya r chaarjii qabxii tokko irraa ta'e barbaadan kana fayyadamaa.

Of eeggadhu

Dirreen chaarjii test q irratti hin hirkatu, kanaaf q irratti hin baaddatinaa. Chaarjii hedduuf dirreewwan akka vektaraatti, akaakuu isaanii wajjin ida'aa, akka guddina qofaatti miti. Dirreen madda mallattoo waliin ni baay'ata, madda mallattoo hin qabne irraa ni xiqqaata.

Grade 11 Physics, fuula 217-286 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Seera OhmFoormulaa

R = \frac{V}{I}

R = \frac{V}{I}

$R$: dhorkii kondaaktaraa ykn dhangaggeessaa $\Omega$
$V$: garaagarummaa dandeettii kondaaktara irra jiru $V$
$I$: yaa'ii kondaaktara keessa darbu $A$

Yoom akka itti fayyadamtu

Yeroo dhorkii, voltii ykn yaa'ii kondaaktara bitaa keessatti yoo ho'ichi kan itti fufu barbaadan kana fayyadamaa.

Of eeggadhu

Seerri Ohm ho'ichi yeroo itti fufu qofaa hojjeta. Ibsaa filaamentii ohmic miti: yeroo ho'u dhorkiin isaa ni guddata, kanaaf garafni V-I sarara qajeelaa miti, dheebotee ta'a.

Grade 11 Physics, fuula 217-286 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Gaaffii shaakalaa boqonnaa 6

Gaaffii filannoo 10 boqonnaa kana irratti barreeffaman, tokkoon tokkoon isaanii deebii fi sababa qabu. Deebii banuu dura yaali. Gaaffiin kun Ingiliffaa tiin.

Gaaffii 1

According to the quantization of electric charge, how many electrons are needed to make a total charge of -1.0 C?

A.6.25 x 10^18 electrons
B.1.60 x 10^19 electrons
C.6.25 x 10^-19 electrons
D.1.60 x 10^-18 electrons

Deebii fi sababa agarsiisi

Deebii: A 6.25 x 10^18 electrons

Sababa

Charge is quantized as q = ne. Dividing the charge 1.0 C by the elementary charge 1.6 x 10^-19 C gives 6.25 x 10¹⁸ electrons.

Gaaffii 2

Which of the following is a correct property of electric field lines?

A.They cross each other at right angles.
B.They begin on positive charges and end on negative charges.
C.They are parallel to the surface of a charged conductor.
D.Farther spaced lines indicate a stronger electric field.

Deebii fi sababa agarsiisi

Deebii: B They begin on positive charges and end on negative charges.

Sababa

Electric field lines always begin on positive charges and radiate outward toward negative charges, where they terminate. They never cross each other and are always perpendicular to the surface of a charged body.

Gaaffii 3

Which of the following pairs of units are both valid units for electric field strength?

A.N/C and V/m
B.J/C and V/m
C.N/m and V/C
D.A/m and N/C

Deebii fi sababa agarsiisi

Deebii: A N/C and V/m

Sababa

Electric field strength is defined as force per unit charge (N/C), which is also mathematically equivalent to potential difference per unit distance (V/m): 1 N/C = 1 V/m.

Gaaffii 4

A headlamp in a vehicle draws a current of 2.50 A when a potential difference of 12.0 V is applied across it. What is its resistance?

A.0.21 ohms
B.4.8 ohms
C.14.5 ohms
D.30.0 ohms

Deebii fi sababa agarsiisi

Deebii: B 4.8 ohms

Sababa

Using Ohm's law, R = V / I. Substituting the values gives R = 12.0 V / 2.50 A = 4.8 ohms.

Gaaffii 5

Two point charges of +2.0 nC and +8.0 nC are separated in air by a distance of 0.20 m. What is the magnitude of the electrostatic force between them? (Take k = 9.0 x 10⁹ N m²/C²)

A.3.6 x 10^-6 N
B.7.2 x 10^-6 N
C.3.6 x 10^-4 N
D.1.8 x 10^-5 N

Deebii fi sababa agarsiisi

Deebii: A 3.6 x 10^-6 N

Sababa

By Coulomb's law: F = k (q₁ q₂) / r². Substituting the numbers: F = (9.0 x 10⁹ x 2.0 x 10^-9 x 8.0 x 10^-9) / (0.20)² = 1.44 x 10^-7 / 0.04 = 3.6 x 10^-6 N.

Gaaffii 6

What is the electric potential produced by a point charge of +1.0 μC at a distance of 1.0 mm from it? (Take k = 9.0 x 10⁹ N m²/C²)

A.9.0 kV
B.90 kV
C.9.0 MV
D.90 MV

Deebii fi sababa agarsiisi

Deebii: C 9.0 MV

Sababa

Using V = k Q / r: with Q = 1.0 x 10^-6 C and r = 1.0 x 10^-3 m, we get V = (9.0 x 10⁹ x 1.0 x 10^-6) / (1.0 x 10^-3) = 9.0 x 10⁶ V = 9.0 MV.

Gaaffii 7

Which of the following is true concerning equipotential lines and electric field lines?

A.Equipotential lines can intersect each other.
B.Work is required to move a test charge along an equipotential line.
C.Equipotential lines are always parallel to electric field lines.
D.Equipotential lines are always perpendicular to electric field lines.

Deebii fi sababa agarsiisi

Deebii: D Equipotential lines are always perpendicular to electric field lines.

Sababa

Equipotential lines are always perpendicular to electric field lines. Since the potential is constant along an equipotential line, no work is required to move a charge along it, and two equipotential surfaces never intersect.

Gaaffii 8

A 4.0 ohm resistor and a 12.0 ohm resistor are connected in parallel. This combination is connected in series with a 5.0 ohm resistor and an 8.0 ohm resistor. What is the total equivalent resistance of the circuit?

A.8.0 ohms
B.16.0 ohms
C.24.0 ohms
D.29.0 ohms

Deebii fi sababa agarsiisi

Deebii: B 16.0 ohms

Sababa

First, find the parallel resistance: R_p = (4.0 x 12.0) / (4.0 + 12.0) = 48 / 16 = 3.0 ohms. Then add the series resistors: R_eq = 5.0 + 3.0 + 8.0 = 16.0 ohms.

Gaaffii 9

A flat square surface of area 2.0 cm by 2.0 cm lies in the xy-plane. An electric field is given by E = (20 j + 50 k) N/C. What is the electric flux through this square surface?

A.0.008 N m^2/C
B.0.02 N m^2/C
C.0.20 N m^2/C
D.2.8 N m^2/C

Deebii fi sababa agarsiisi

Deebii: B 0.02 N m^2/C

Sababa

Because the surface is in the xy-plane, its area vector points along the z-axis (k-direction): A = (0.02 m x 0.02 m) k = 4 x 10^-4 m² k. The electric flux is Φ = E • A = (50)(4 x 10^-4) = 0.02 N m²/C.

Gaaffii 10

A galvanometer has a coil resistance of 50 ohms and gives full-scale deflection with a current of 5 mA. What shunt resistance must be connected in parallel to convert it into an ammeter with a full-scale range of 10 A?

A.0.025 ohms
B.0.25 ohms
C.2.5 ohms
D.50.0 ohms

Deebii fi sababa agarsiisi

Deebii: A 0.025 ohms

Sababa

The required shunt resistance is R_sh = [I_g / (I - I_g)] x R_g. Substituting the given values: R_sh = [0.005 / (10 - 0.005)] x 50 = (0.005 / 9.995) x 50 ≈ 0.025 ohms.

Appiin boqonnaa kana irratti gaaffii 12 qaba; utuu hojjettuu ni sirreessa. Kan asii jiran 10 jalqabaa ti.

Fuula kitaabaa boqonnaa 6

Boqonnaan 6 Maxxansa Ministeera Barnootaa keessatti fuula 217 hanga fuula 286 ti.

Maxxansa Ministeera Barnootaa, Ingiliffaa, Fuula 338

PDF buufadhaa

Yaadannoo gabaabaa Fiizikii kutaa 11 hunda

Boqonnaa kana appii keessatti shaakali

Temari boqonnaa kana gara shaakala yeroon daangeffametti jijjiira; utuu hojjettuu ni sirreessa; wanti si hin galinis barsiisaa AI tiin ni ibsama.

Bilisaan jalqabi

Temari.et appii Android qaba; Google Play irratti bilisaan argama.

Waa'ee appii Temari