The distance it traveled is 147
The answer is , Conductance
Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;
where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:
The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:
Total potential due to this charges = 4500 V + 6750 V = 11250 V
Answer:
dB = (-5 × 10⁻⁷k) T
Magnitude of dB = (-5 × 10⁻⁷) T; magnitude is actually (5 × 10⁻⁷) T in the mathematical sense of what magnitude is, but this question instructs to include the negative of the direction of dB is negative.
Direction is in the negative z-direction as evident from the sign on dB's vector notation.
Explanation:
From Biot Savart's relation, the magnetic field is given by
dB = (μ₀I/4πr³) (dL × r)
μ₀ = (4π × 10⁻⁷) H/m
I = 5.0 A
r = (2î) m
Magnitude of r = 2
dL = (4j)
(dL × r) is the vector product of both length of current carrying wire vector and the vector position of the point where magnetic field at that point is needed.
(dL × r) = (4j) × (2î)
|i j k|
|0 4 0|
|2 0 0|
(dL × r) = (0î + 0j - 8k) = (-8k)
(μ₀I/4πr³) = (4π × 10⁻⁷ × 5)/(4π×2³)
(μ₀I/4πr³) = (6.25 × 10⁻⁸)
dB = (μ₀I/4πr³) (dL × r) = (6.25 × 10⁻⁸) × (-8k)
dB = (-5 × 10⁻⁷k) T
Magnitude of dB = (-5 × 10⁻⁷) T; magnitude is actually (5 × 10⁻⁷) T in the mathematical sense of what magnitude is, but this question instructs to include the negative of the direction of dB is negative.
Direction is in the negative z-direction as evident from the sign on dB's vector notation.
Hope this Helps!!!
