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mars1129 [50]
3 years ago
9

Lionel makes a graphic organizer to compare the electric field around a positive charge with the electric field around a negativ

e charge.
Which labels belong in regions X, Y, and Z?


X: Points away from the charge

Y: Points in the direction of the force on the positive charge

Z: Points toward the charge

X: Points away from the charge

Y: Points in the direction of the force on the negative charge

Z: Points toward the charge

X: Points toward the charge

Y: Points in the direction of the force on the positive charge

Z: Points away from the charge

X: Points toward the charge

Y: Points in the direction of the force on the negative charge

Z: Points away from the charge
Physics
2 answers:
krek1111 [17]3 years ago
6 0
X- points away from the charge
y- points in the direction of the force on the positive charge
z- points toward the charge
aksik [14]3 years ago
4 0

Answer:

The Answer is A <-------------

Explanation:

Because I took the test on Edge

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Answer:

Part a)

V = -1.52 V

Part b)

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Explanation:

Part a)

Electric potential is a scalar quantity

so here we can say that total potential due to a ring on its center is given as

V = \frac{kQ}{R}

here we know that

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now we have

V = \frac{(9\times 10^9)(-5\times 2.70 \times 10^{-12})}{0.08}

V = -1.52 V

Part b)

Potential on the axis of the ring is given as

V = \frac{kQ}{\sqrt{r^2 + R^2}}

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V = -1.16 V

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Two long straight parallel wires are separated by 7.0cm. There is a 2.0A current flowing in the first wire. If the magnetic fiel
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Answer:

The current in the second wire is 4.4 A.

Explanation:

Given that,

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Current in first wire = 2.0 A

The magnetic field strength is zero at distance of 2.2 cm from the first wire.

We need to calculate the current in the second wire

Using formula of magnetic field

B=B_{1}-B_{2}

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\dfrac{\mu_{0}I_{1}}{2\pi r_{1}}=\dfrac{\mu_{0}I_{2}}{2\pi r_{2}}

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Here, r_{2}=d-r_{1}

I_{2}=\dfrac{I_{1}\times(d-r_{1})}{r_{1}}

Put the value into the formula

I_{2}=\dfrac{2.0\times(7.0-2.2)}{2.2}

I_{2}=4.4\ A

Hence, The current in the second wire is 4.4 A.

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