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Amiraneli [1.4K]
3 years ago
7

Use the principle of superposition (Equation 1.3) to predict the electric field at the point 0.5 meters to the right and 1.5 met

ers above the 5 nC charge. Express your result both in terms of x and y components, and then in terms of total magnitude and direction.
Physics
1 answer:
kvasek [131]3 years ago
4 0

Answer:

Explanation:

The x component of electric field Ex = k Q / R²

= 9 x 10⁹ x 5 x 10⁻⁹ / .5²

= 180 N .

The y component of electric field Ey = k Q / R²

The y component of electric field Ey

= 9 x 10⁹ x 5 x 10⁻⁹ / 1.5²

= 20 N

Total field = \sqrt{180^2+20^2}

= 181.1 N .

Angle with x axis = Tanθ = Ey / Ex = 20 / 180 = .111

= 6.33⁰

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How much voltage is required to run 0.64 A of current through a 240 resistor? Use V= IR.
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B. If you press that into a calculator it comes up with 153.6. You then shift the decimal point 2 times forward and you end up getting 1.5 x 10^2 V.
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At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does
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(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

V = V_R + V_L

The potential difference across the inductance is given by

V_L(t) = V e^{-\frac{t}{\tau}} (1)

where

\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s is the time constant of the circuit

At time t=0,

V_L(0) = V e^0 = V = 20 V

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

V_R = V-V_L = 20 V-20 V=0

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

t >> \tau

Since \tau is at the order of less than milliseconds.

Using eq.(1), we see that for t >> \tau, the exponential becomes zero, and therefore the potential difference across the coil is zero:

V_L = 0

Therefore, the potential difference across the resistor will be

V_R = V-V_L = 20 V- 0 = 20 V

(c) Yes

The two voltages will be equal when:

V_L = V_R (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

V=V_L +V_R

And rewriting this equation,

V_R = V-V_L

Substituting into (2) we find

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So, the two voltages will be equal when they are both equal to 10 V.

(d) at t=5.77\cdot 10^{-4}s

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V_L=\frac{V}{2}

Using eq.(1), and this last equation, this means

V e^{-\frac{t}{\tau}} = \frac{V}{2}

And solving the equation for t, we find the time t at which the two voltages are equal:

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I_0 = 3.20 A

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V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

V_L(0)=.V

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

t>>\tau

If we use this approximation into the formula

V_L(t) = -V e^{-\frac{t}{\tau}} (1)

We find that

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Answer:

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