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dalvyx [7]
3 years ago
11

An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance

R away from the wire using Gauss's Law for Electrostatics.
Physics
1 answer:
marshall27 [118]3 years ago
4 0

Answer:

E = \frac{\lambda}{2\pi \epsilon_0 r}

Explanation:

Let the linear charge density of the charged wire is given as

\frac{q}{L} = \lambda

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

\int E. dA = \frac{q}{\epsilon_0}

here we have

E \int dA = \frac{\lambda L}{\epsilon_0}

E. 2\pi r L = \frac{\lambda L}{\epsilon_0}

so we have

E = \frac{\lambda}{2\pi \epsilon_0 r}

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The moon is 3.85 x 10 to the 8 m from earth and has a diameter of 3.48 x 10 to the 6 m. You have a pea (diameter = 0.50 cm) and
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Answer:

(a) dime

Explanation:

Convert all to metric unit:

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1.8 cm = 0.018 m

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In order to find out we would need to calculate the ratio R between the object diameter d and their distance s to our eyes:

R_m = \frac{d_m}{s_m} = \frac{3.48*10^6}{3.85*10^8} \approx 0.009

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R_d = \frac{d_d}{s_d} = \frac{0.018}{0.71} \approx 0.0253

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A ball is thrown up into the air with an initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed
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Answer:

B) t = 1.83 [s]

A) y = 16.51 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f} =v_{o} -g*t

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Vf = final velocity = 0

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g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.

A) The maximum height is reached when the final velocity of the ball is zero.

0 = 18 - (9.81*t)

9.81*t = 18

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