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>
Work done = force * distance
2m * 9.8
work done = 19.6 J
Explanation:
When the wire is connected to a battery, the compass needle moves and changes its position. This happens because the needle magnetizes the copper wire, thus, creating a force.
While the current in the wire produces a magnetic field and exerts a force on the needle. The insulation on the wire becomes energized and exerts a force on the needle. Hence, the compass needle moves and changes its position.
Answer:
16.405m/s
Explanation:
Given, initial velocity = u = 1.93m/s, acceleration = a = 0.750m/s2, time = t = 19.3s, final velocity = v= ?
Using the first equation of linear motion,
v = u + at
v = 1.93 + 0.750 x 19.3
v = 1.93 + 14.475
v = 16.405m/s
Answer:
a. 475.14 Hz
b. 1959 Hz
c. 2341.53 Hz , 3053.34 Hz
Explanation:
a. smallest use the capacitive 4.2 uF + 6.0 uF = 10.2uF replacing:
b. second smallest use the capacitive 6 uF so:
c. second largest and largest oscillation first combination so:
Use 4.2 uF
And finally largest oscillation cap in serie so:
