Calculate potential energy
1 answer:
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Answer:
Magnitude of the resulting force on the 7 nC charge at the origin:
Fn₁= 23.95*10⁻⁹ N
Explanation:
Look at the attached graphic:
Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.
q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:
F = (k*q₁*q)/(d)²
m : distance from q₁ to q₂
(d₁₂)² = 18 m²
m : distance from q₁ to q₃
(d₁₃)² = 10 m²
m : distance from q₁ to q₄
(d₁₄)² = 13 m²
K= 8.98755 × 10⁹ N *m²/C²
q₁= 7*10⁻⁹C
k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9
F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C
F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C
F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C
x-y components of the net force on q₁ (Fn₁):
α= tan⁻¹(3/3)= 45° , β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°
Fn₁x = F₂₁x+ F₃₁x+F₄₁x
F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N
F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N
F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N
Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N
Fn₁x = -23.87 *10⁻⁹ N
Fn₁y = F₂₁y+ F₃₁y+F₄₁y
F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N
F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N
F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N
Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N
Fn₁y = 1.97*10⁻⁹ N
Magnitude of the resulting force on the 7 nC charge at the origin (q₁):
Fn₁= 23.95*10⁻⁹ N
.
The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:
where,
the total power emitted,
the surface area of the body,
the Stefan-Boltzmann Constant, and
the temperature of the body in degrees Kelvins.
.
.
.
Keep as many significant figures in as possible. The error will be large when
is raised to the power of four. Also, the real value will be much smaller than
since the emittance of a human body is much smaller than assumed.