Electric field due to a charged rod along its axis is given by

here we know that
L = 14 cm
r = distance from end of rod
r = 36 - 7 = 29 cm
Q = 222 mC
now we will have


Answer:
The charge on the outer surface of the block = -5.00 nC
The charge on the surface of the cavity (on the inner surface of the block) = -3.00 nC
Explanation:
The point charge within the cavity will induce a charge equal in magnitude and opposite in sign on the inside cavities of the copper block.
Charge of the point charge = 3.00 nC
Charge induced on the inner surface of the Copper block's cavity = -3.00 nC
Since the charge on a conductor should usually be neutral, the charge on the inner surface causes a charge equal in magnitude and also opposite in sign on the outer surface of the block; that is, 3.00 nC.
But this block already has an excess charge of -8.00 nC (which resides on the surface because excess charge for conductors reside on the surface of the conductors)
So, net charge on the outer surface of the Copper block = -8.00 + 3.00 = -5.00 nC.
Hope this Helps!!!
In other words a infinitesimal segment dV caries the charge
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
Answer:
1: surface temperature
2: red giant
3: The brightest stars are called supergiants. Star clusters are rich in stars just off the main sequence called red giants. Main sequence stars are called dwarfs.
4: A white dwarf is very dense
5: red giant
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Molecular formula of water molecule is H₂O.