Lub zog fais fab - cov teeb meem thiab kev daws teeb meem

1. Ib lub electron raug nrawm los ntawm qhov so los ntawm qhov sib txawv ntawm 12 V. Qhov kev hloov pauv ntawm lub zog fais fab ntawm lub electron yog dab tsi?

Paub:

Tus nqi ntawm ib lub electron (e) = -1.60 x 10 ^-19 Coulomb

Lub zog hluav taws xob = voltage (V) = 12 Volts

Xav tau: Kev hloov pauv ntawm lub zog hluav taws xob ntawm lub electron (ΔPE)

Tshuaj:

ΔPE = q V = (-1.60 x 10 ^-19 C)(12 V) = -19.2 x 10 ^-19 Joules

Lub cim rho tawm qhia tias lub zog muaj peev xwm txo qis.

[irp]

2. Ob lub phaj sib luag raug them nqi. Qhov sib nrug ntawm cov phaj yog 2 cm thiab qhov loj ntawm lub zog hluav taws xob ntawm cov phaj yog 500 Volt / 'meter'. Qhov kev hloov pauv ntawm lub zog muaj peev xwm ntawm proton thaum raug nrawm los ntawm lub phaj them nqi zoo mus rau lub phaj them nqi tsis zoo yog dab tsi.

Paub:

Qhov loj ntawm lub zog hluav taws xob ntawm cov phaj (E) = 500 Volt/meter

Qhov nrug ntawm cov phaj (s) = 2 cm = 0,02 m

Tus nqi ntawm ib qho proton = +1.60 x 10 ^-19 Coulombs

Xav tau: Kev hloov pauv ntawm lub zog hluav taws xob (ΔPE)

Tshuaj:

Lub zog hluav taws xob:

V = E s

V = (500 Volt/m2)(0.02 m)

V = 10 Volts

Kev hloov pauv ntawm lub zog fais fab:

ΔPE = qV

ΔPE = (1,60 x 10 ^-19 C)(10 V)

ΔPE = 16 x 10 ^-19 Joule

ΔPE = 1.6 x 10 ^{-1 8} Joule

[irp]

3. Ob lub ntsiab lus them nqi raug sib cais los ntawm qhov deb ntawm 10 cm. Them rau ntawm qhov chaw A = +9 μC thiab them rau ntawm qhov chaw B = -4 μC. k = 9 x 10 ⁹ Nm ² C ⁻² , 1 μC = 10 ⁻⁶ C. Qhov kev hloov pauv ntawm lub zog fais fab ntawm qhov chaw B yog dab tsi yog tias raug nrawm mus rau qhov chaw A?

Paub:

Them A (q ₁ ) = +9 μC = +9 x 10 ⁻⁶ C

Them B (q ₁ ) = -4 μC = -4 x 10 ⁻⁶ C

k = 9 x 10 ⁹ Nm ² C ⁻²

Qhov nrug ntawm cov nqi A thiab B (r) = 10 cm = 0.1 m = 10 ^-1 m

Xav tau: Kev hloov pauv ntawm lub zog hluav taws xob (ΔEP)

Tshuaj: