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True [87]
3 years ago
12

A thin spherical shell of radius 15 cm carries 4.8 μC, distributed uniformly over its surface. At the center of the shell is a p

oint charge. Part A If the electric field at (just barely outside) the surface of the sphere is 750 kN/C and points outward, what is the charge of the point charge?
Physics
1 answer:
zvonat [6]3 years ago
8 0

Explanation:

It is known that on the surface of the shell the electric field will be the sum of electric field due to the shell and due to the point charge.

          E_{shell} + E_{p.cha} = 750 \times 10^{3} N/C

    \frac{kq_{shell}}{r^{2}} + \frac{kq_{p.ch}}{r^{2}} = 750 \times 10^{3} N/C

    (q_{shell} + q_{p.cha}) = \frac{750 \times 10^{3} N/C}r^{2}{k}

    4.8 \times 10^{-6} C + q_{p.cha} = \frac{(750 \times 10^{3}N/C)(15 \times 10^{-2}m)^{2}}{9 \times 10^{9} Nm^{2}/C^{2}}

     q_{p.cha} = -2.93 \times 10^{-6} C (\frac{1.0 \muC}{10^{-6}C})

                = -3.0 \muC

Thus, we can conclude that charge of the point charge is -3.0 \muC.

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<em>Measure the amount of water it displaces.</em>

This won't be easy, because the bat floats in water.  But I think you can get around that little problem like this:

-- Get some kind of a tank or tub that's big enough to hold the whole bat under water.

-- Get a heavy weight, like a big wrench or a small rock.  

-- Fill the tub almost to the tippy top with water.

-- Slip the heavy weight into the tub, slowly.  Some water will run over the top and out of the tub.  That's OK ... it's exactly what you want.  If NO water runs over the top, pour some more in, until it runs out and then stops.  You want the tub full to the brimmy rim with the rock at the bottom of it.

-- Take the heavy weight out of the tub.

-- Now set the tub into a bigger tub or a deep pan.  The next time it overflows and some water runs out of it, you'll need to catch that water and measure it.

-- Get a short piece of heavy string.  Tie the heavy weight to somewhere near the middle of the bat.

-- Slowly slide the bat into the water, with the rock tied to it.  The bat needs to go complete underwater.

-- Some more water will run over the top and out of the tub, and INTO the lower tub.  Wait until the overflow stops and everything settles down again.

-- Take the bat (tied to the weight) out of the tub.  Slowly and carefully, so that your hand or your arm doesn't make any MORE water run over and out.

-- Lift the upper tub out of the lower tub.

-- Take the lower tub, with the overflow water in it.  Using a kitchen measuring cup, or a saucepan or a bottle, or anything else with liquid amounts marked on it, measure how much water overflowed into the lower tub.

THAT amount is the volume of the bat.

You may have to do some units conversions.  Like if you need the volume of the bat in cm³ and you used measuring vessels marked in fluid ounces.  But you can find all those conversion factors with a search on Floogle.

8 0
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Answer:

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See attached file

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Answer:

E = 401.3 eV

Explanation:

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so we will have

E = \frac{hc}{\lambda}

here we have

hc = 1240 nm-eV}

also we have

\lambda = 3.09 nm

so we will have

E = \frac{1240}{3.09} eV

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I think that when work is done and a force istransferred an object must move in the direction of the force.
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The difference is P1-P2 = 1.1-(-0.88)=1.98kg m/s
4 0
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