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madreJ [45]
3 years ago
6

What is the electric force acting between two charges of -0.0045 C and -0.0025 C that are 0.0060 m apart? Use Fe=kq1q2/r^2 and k

= 9.00 x 10^9 N*m^2/C^2
A. 1.7 x 10^7 N
B. -1.7 x 10^7 N
C. -2.8 x 10^9 N
D. 2.8 x 10^9 N
Physics
2 answers:
oksano4ka [1.4K]3 years ago
4 0

Answer:

D. 2.8 × 10⁹ N

Explanation:

The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.

Fe= k Q₁Q₂/r²

Q₁= -0.0045 C

Q₂= -0.0025 C

r= 0.0060 m

k= 9.00 × 10 ⁹ Nm²/C²

Fe= (9.00 × 10 ⁹ Nm²/C²×-0.0045 C×-0.0025 C)/0.0060²

=2.8 × 10⁹ N

ArbitrLikvidat [17]3 years ago
3 0

Answer:

Like the other person said, the answer is 2.8 x 10^9

Explanation:

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In this experiment we will observe the magnetic fields produced by a current carrying wire. A long wire is suspended vertically,
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See explanation

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Solution:-

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( See attachments )

- The equation for the magnetic field strength - B - (magnitude) produced by a long straight current-carrying wire is given by the Biot Savart Law:

                                  B = \frac{uo*I}{2\pi *r}

Where,

I : The current,

r : The shortest distance to the wire,

uo : The permeability of free space. = 4π * 10^-7  T. m/A

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                             B = \frac{uo*I}{4\pi }*\int\frac{dl xr}{r^2}      

Where the integral sums over,

 1) The wire length where vector dl = direction of current (in or out of plane)

 2) r is the distance between the location of dl and the location at which the magnetic field is being calculated

 3)  r^ is a unit vector in the direction of r.

   

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3 years ago
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