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Sati [7]
3 years ago
5

A test charge of -1.4 × 10-7 coulombs experiences a force of 5.4 × 10-1 newtons. Calculate the magnitude of the electric field c

reated by the negative test charge.
A. 1.4 × 106 newtons/coulomb
B. 1.9 × 106 newtons/coulomb
C. 5.4 × 10-1 newtons/coulomb
D. 3.6 × 106 newtons/coulomb
Physics
2 answers:
kotykmax [81]3 years ago
8 0

Answer: D. 3.6*10^-6

Explanation:

Electric field strength E= Force on a charge / charge

E = F/q

E = 5.4*10^-1 / 1.4*10^-7

E = 3.857*10^6 N/C

mr_godi [17]3 years ago
3 0
I would say C 5.4 x 10-1 Newtons/coulomb
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Explanation:

21)

The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:

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A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

5 0
3 years ago
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