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11Alexandr11 [23.1K]
3 years ago
8

The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radi

us of the uranium nucleus is approximately 7.4×10−15m .
a. What is the electric field this nucleus produces just outside its surface?Express your answer using two significant figures.
b. What magnitude of electric field does it produce at the distance of the electrons, which is about 1.1×10-10m ?
c. The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?Express your answer using two significant figures.=____________N/C
Physics
1 answer:
guapka [62]3 years ago
7 0

Answer:

(A) E = 2.633 * 10^19 N/C

(B) E = 1.096 * 10^13 N/C

(C) E = -1.096 * 10^13 N/C

Explanation:

Parameters given:

Number of protons, N = 92

Radius of Uranium nucleus = 7.4 * 10^-15 m

Electronic charge, ẹ = 1.6023 * 10^-19 C

Electric field at a point R due to a charge Q is given as

E = (k*Q) / (R^2)

Where k =Coulombs constant

(A) Since there are 92protons,the total Electric field due to the protons will be:

E = (k*e*N) / (R * R)

E = (9 * 10^9 * 92 * 1.6023 * 10^-19) / (7.4 * 10^-15)^2

E = 2.633 * 10^19 N/C

(B) At the position of the electrons, R = 1.1 * 10^-10m. Therefore, Electric field will be:

E = (9 * 10^9 * 92 * 1.6023 * 10^-19) / (1.1 * 10^-10)^2

E = 1.096 * 10^13 N/C

(C) There are 92 electrons in the Uranium atom and electrons have a charge - e, hence, the Electric field due to the electrons at the nucleus will be:

E = -(k*e*N) / (R * R)

E = -(9 * 10^9 * 92 * 1.6023 * 10^-19) / (1.1 * 10^-10)^2

E = -1.096 * 10^13 N/C

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This problems a perfect application for this acceleration formula:

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During the speeding-up half:     1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half:    1,600 meters = (1/2) (1.3 m/s²) T²

Pick either half, and divide each side by  0.65 m/s²: 

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I'm completely lost in physics, the Kinematics and dynamics units.. My exam is coming up.. If a biker travelling at 6.4m/s sees
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D=rt
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