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3 years ago

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1. In an atom, an electron and a proton are separated by a distance of 1 x 10-8m. a) Find the force on the electron. b) Assuming

a circular orbit (which wouldn't happen, but we'll go with it). Find the velocity of the electron.

1 answer:

aleksandrvk [35]3 years ago

6 0

Answer:

(a) 2.3 x 10^-12 N

(b) 1.6 x 10^5 m/s

Explanation:

q1 = 1.6 x 10^-19 C

q2 = 1.6 x 10^-19 C

r = 1 x 10^-8 m

(a) the electrostatic force is given by

F = K q1 x q2 / r^2

F = ( 9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (1 x 10-^-8)^2

F = 2.3 x 10^-12 N

The force is attractive in nature because the nature of charge on electron and proton s opposite to each other.

(b) The electrostatic force is balanced by the centripetal force actig on the electron.

F = mv^2 / r

where, r be the radius of orbit and v be the velocity of electron in the orbit.

2.3 x 10^-12 = (9.1 x 10^-31 x v^2) / (1 x 10^-8)

v = 1.6 x 10^5 m/s

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A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

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Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

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$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

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To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

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$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

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$k * R^2= \sigma \times R^2$

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Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

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The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

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$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

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