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Pepsi [2]
3 years ago
8

Two small spheres spaced 20.0 cm apart have equal charge. how many excess electrons must be present on each sphere if the magnit

ude of the force of repulsion between them is 4.57 * 10-21 n?
Physics
1 answer:
uysha [10]3 years ago
8 0
<span>890 Coulomb's law is F = k * q1 * q2 / r^2 where F = force k = Coulomb's constant 8.99x10^9 N m^2/C^2 q1,q2 = signed charges r = distance between charges Since the charges are the same, let's simplify the equation, solve for q, then substitute the known values and calculate. F = k * q1 * q2 / r^2 F = k * q^2 / r^2 F*r^2 = k * q^2 F*r^2/k = q^2 sqrt(F*r^2/k) = q sqrt(4.57x10^-21 N * (0.2m)^2 / (8.99x10^9 N m^2/C^2)) = q sqrt((4.57x10^-21 N * 0.04m^2) / (8.99x10^9 N m^2/C^2)) = q sqrt((1.828x10^-22 N*m^2) / (8.99x10^9 N m^2/C^2)) = q sqrt(2.03337x10^-32 C^2) = q 1.42596x10^-16 C = q So each sphere has to have an excess of 1.42596x10^-16 Coulombs of electrons. A coulomb is 6.24150934x10^18 electrons, so let's do the multiplication: 1.42596x10^-16 * 6.24150934x10^18 = 8.9001426584664x10^2 = 890 So each sphere has an extra 890 electrons.</span>
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Answer:

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b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

<u>Given :</u>

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( K_{A}= 0) to the point B at distance l = 0.500m where its kinetic energy is (  K_{B}= 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

<u>Required :</u>

<em>(a) We are asked to find the electric potential VB </em>

<em>(b) We want to determine the magnitude and the direction of the electric field E. </em>

<u> Solution </u>

(a) We are given the values for VA,K_{B} and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

                                   K_{A} +U_{A} =K_{B} +U_{B} .........................................(1)                                          

Where K_{A}= 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential.  So, equation (1) will be in the form :

                                  0+qVA=K_{B} +qVB                      (Divide by q)

                                         VA=K_{B} /q + VB                  (solve for VB)

                                         VB=VA- K_{B}/q .......................................(2)

We get the relation between VB, VA and K_{B}, now we can plug our values for VA, K_{B} and q into equation (2) to get VB

                                         VB=VA- K_{B}/q

                                              =30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

                                              =80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

                                   VA-VB =\int\limits^1_0 {E} \, dl...................................(a)

                                               =E\int\limits^1_0 {} \, dl

                                   VA-VB=El                      (solve for E)

                                            E= VA-VB/l..................................(3)

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The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

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