A)<span>
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² -
r³/R) dr
which when integrated from 0 to r is
total charge = 4π * ρ0 (r³/3 + r^4/(4R))
and when r = R our total charge is
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3
and after substituting ρ0 = 3Q / πR³ we have
total charge = Q ◄
B) E = kQ/d²
since the distribution is symmetric spherically
C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 -
r/R)dr
so
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is
and after substituting for ρ0 is
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4))
which could be expressed other ways.
D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that
r = R for a min/max (and we know it's a max since r = 0 is a
min).
<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>
Start with saying “If...” and then say “the the....”
False, 1000 meters is a kilometer, so 1 kilometer is 1000 meters, meaning to convert meters to kilometers, you divide by 1000, making it false.
Answer:
The power dissipated in either one of the parallel resistors is 2 V
Explanation:
Given;
two parallel resistors, R₁ and R₂ = 2 ohms
The total resistance of the Two resistors of 2 ohms connected in parallel is;
when connected to another resistor of 1 ohm in series, the total resistance becomes;
Rt = R₁ + R₂
Rt = 1 + 1 = 2 ohms
Current in the circuit, I = voltage / total resistance
= 2 /2 = 1 A
the overall circuit has been resolved to series connection, and current flow in series circuit is constant.
Power = I²R
Thus, power dissipated in either one of the parallel 2 ohms resistors is;
Power = I²R = (1)² x 2 = 2 V
The Answer is= 7.8 x 10^4