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son4ous [18]
2 years ago
9

Uma partícula com carga “Q”, no vácuo, gera um potencial elétrico de 600 volts a uma distância de 6,0 m, determine o valor dessa

carga. Em seguida, assinale a alternativa que apresenta o valor dessa carga.

Physics
1 answer:
garik1379 [7]2 years ago
8 0

Answer:

q =  400 nC

the correct answer is b

Explanation:

The expression for the electric potential of a point charge is

            V = k q / r

they ask us for the electrical charge

          q = V r / k

let's calculate

         Q = 600 6.0 / 9 10⁹

         Q = 4 10⁻⁷ C

let's reduce to nC

          Q = 4 10⁻⁷ C (10⁹ nC / 1C)

          q = 4 10² nC = 400 nC

the correct answer is b

Traslate

La expresión para el potencial eléctrico de una carga puntual es

            V = k q/r

nos piden la carga eléctrica

          q= V r /k

          calculemos

         Q= 600  6,0 / 9 10⁹

         Q=  4 10⁻⁷ C

reduzcamos a nC

          Q = 4 10⁻⁷ C(10⁹ nC/1C )  

          q = 4 10² nC = 400 nC

la respuesta correcta es b

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A solenoid coil with 22 turns of wire is wound tightly around another coil with 340 turns. The inner solenoid is 25.0 cm long an
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Answer:

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b) 1.18*10^-5 H

c) 20mV

Explanation:

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\Phi_B=BA=\mu_oNIA

mu_o: magnetic permeability of vacuum = 4pi*10^-7 T/A

N: turns of the solenoid = 340

I: current of the inner solenoid = 0.100A

A: area of the inner solenoid = pi*r^2

r: radius of the inner solenoid = 2.00cm/2=1.00cm=10^-2m

You calculate the area and then replace the values of N, I, mu_o and A to find the magnetic flux:

A=\pi(10^{-2}m)^2=3.141510^{-4}m^2\\\Phi_B=(4\pi*10^{-7}T/A)(340)(0.100A)(3.1415*10^{-4}m^2)=1.34*10^{-8}W\\

the magnetic flux is 1.34*10^{-8}W

b) the mutual inductance is given by:

M=\mu_o N_1 N_2 \frac{A_2}{l}

N1: turns of the outer solenoid = 22

N2: turns of the inner solenoid

A_2: area of the inner solenoid

l: length of the solenoids = 25.0cm=0.25m

by replacing all these values you obtain:

M=(4\pi*10^{-7}T/A)(340)(22)\frac{3.14*10^{-4}m^2}{0.25m}=1.18*10^{-5}H

the mutual inductance is 1.18*10^{-5}H

c) the emf induced can be computed by using the mutual inductance and the change in the current of the inner solenoid:

\epsilon_1=M\frac{dI_2}{dt}

by replacing you obtain:

\epsilon_1=(1.18*10^{-5}H)(1700A/s)=0.02V=20mV

the emf is 20mV

7 0
3 years ago
Read 2 more answers
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