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goldfiish [28.3K]
3 years ago
6

A "gauge 8" jumper cable has a diameter d of 0.326 centimeters. The cable carries a current I of 30.0 amperes. The electric fiel

d E in the cable is 0.062 newtons per coulomb. part a: What electric field E′ would have been required to create a current of 30.0 amperes in a copper "gauge 10" wire with diameter d equal to 0.259 centimeters?
Physics
1 answer:
AveGali [126]3 years ago
4 0

Answer:

0.0979 N/c

Explanation:

Electric field, E is given as a product of resistivity and current density

E=jP where P is resistivity and j is current density

But the current density is given as

j=\frac {I}{A} where I is current and A is area and A=\pi r^{2}

Substituting this into the first equation then E=P\times \frac {I}{\pi r^{2}}

Given diameter of 0.259 cm= 0.00259 m and the radius will be half of it which is 0.001295 m

E=1.72\times 10^{-8}\times \frac {30}{\pi \times 0.001295^{2}}=9.79\times 10^{-2} N/c=0.0979 N/c

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3 years ago
A cubical Gaussian surface surrounds two positive charges, each has a charge q 1 1 = + 3.90 × 10 − 12 3.90×10−12 C, and three ne
Masteriza [31]

Answer:

The electric flux is zero because charge is zero.

Explanation:

Given that,

Positive charge q_{1}=3.90\times10^{-12}\ C

Negative charge q_{2}=-2.60\times10^{-12}\ C

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Using formula of charge

Q_{enc}=2q_{1}+3q_{2}

Put the value into the formula

Q_{enc}=2\times3.90\times10^{-12}+3\times(-2.60\times10^{-12})

Q_{enc}=0

We need to calculate the electric flux

Using formula of electric flux

\phi=\dfrac{Q_{enc}}{\epsilon_{0}}

Put the value into the formula

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7 0
3 years ago
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Answer:

1. 2.5s

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In a certain region of space, the electric potential is V(x,y,z)=Axy-Bx^2+Cy , where A,B , and C are positive constants.a) Calcu
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Answer:

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Explanation:

The electric field and the electric power are related

                    E = - dV / ds

a) Let's find the electric field on the x axis

                  Eₓ = - dV / dx

                  dV / dx = A y - B 2x

                  Eₓ = - A y + 2B x

b) calculate the electric field on the y-axis

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c) the electric field on the z axis

              dv / dz = 0

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.d) at which point the electric field is zero

Since the electric field is a vector quantity all components must be zero

X axis

              0 = = - A y + 2B x

              y = 2B / A x

Axis y

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              .x = -C / A

We substitute this value in the previous equation

             .y = 2B / A (-C / A)

             .y = 2 B C / A2

The correct answer is 3

6 0
3 years ago
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