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erik [133]
3 years ago
7

A thin ring of charge of radius R lies in the x-y plane and is centered on the z-axis. The linear charge density, l, of the loop

depends on the in-plane angle with respect to the x-axis, f. It is given by ______________.
Physics
1 answer:
Kruka [31]3 years ago
7 0

Answer:

Linear charge density (I) = Q/2πR

Explanation:

Linear charge density (I) = charge (Q) per unit length(L)

I = Q/L

For a thin ring of charge with radius R, the length will be equal to the circumference of a circle.

Circumference of a circle = 2πR

Then, the length of the thin ring of charge is 2πR

Linear charge density (I) = Q/2πR

Therefore, for a thin ring of charge of radius R, which lies in the x-y plane and is centered on the z-axis. The linear charge density, l, of the loop is given by Q/2πR

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(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

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7 0
3 years ago
Question 30
stellarik [79]

Answer: 0.69\°

Explanation:

The angular diameter \delta of a spherical object is given by the following formula:

\delta=2 sin^{-1}(\frac{d}{2D})

Where:

d=16 m is the actual diameter

D=1338 m is the distance to the spherical object

Hence:

\delta=2 sin^{-1}(\frac{16 m}{2(1338 m)})

\delta=0.685\° \approx 0.69\° This is the angular diameter

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A track star runs 100 meters in 10 seconds. What is the star's average speed?
kakasveta [241]

Answer:

10m/s

Explanation:

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2 years ago
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