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lidiya [134]
2 years ago
12

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin

um plates, each 21 cm in diameter, separated by 1.0 cm . Part A How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume that value of 3× 10 6 N/C of the field causes a spark. Express your answer with the appropriate units. q max q m a x = nothing nothing SubmitRequest Answer Provide Feedback Next
Physics
1 answer:
Alex777 [14]2 years ago
7 0

Answer:

0.92 μC

Explanation:

In a parallel-plate capacitor, the electric field formed is equal to the charge density divited by the vacuum permisivity e0, as there are no dielectric between the plates. e0 is equal to 8.85*10^-12 C^2/Nm^2. The charge density is the total charge of each individual plate divided by its area. Then, the maximum charge allowed will be equal to:

E = \frac{o}{e_0} = \frac{Q}{Ae_0} \\ Q = E*A*e_0 = 3*10^6 N/C * (0.25*\pi *(0.21m)^2)*8.85*10^{-12}C^2/Nm^2 = 9.196 *10^{-7} C

or 0.92 μC

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Answer:

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Explanation:

From the question we are told that

  From the question we are told that

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         The velocity at t is  v = 16 m/s

  Generally average  angular acceleration is mathematically represented as

                \alpha  = \frac{w_f - w_o}{t}

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            w_f = \frac{v}{l}

                  w_f = \frac{16}{0.85}

                        = 18.823 rad/s

 and w_o is the initial angular velocity which is zero since initial linear velocity is zero

               So

                         \alpha  = \frac{18.823 - 0}{0.15}

                               \alpha =125.487 rad /s^2

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Three small balls of the same size but different masses are hung side-by-side in parallel on the strings of same length. They to
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Answer:

m1/6 ( c )

Explanation:

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m1v = m1v1 + m2v2 --- ( 1 )

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