Use the following equation to help you answer the question. The peak intensity of radiation from a star named Sigma is 2 x 10 6
2 answers:
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Answer:
Explanation:
Use the Kinematic Equation:
Plug in what is given and solve
Answer:
the electric field at point A is
E = 5.5 ×10¹³N/C(-x direction)
Explanation:
given
electrostatics constant k = 9.0×10⁹
charge q = 31.7mC= 31.7×10⁻³C
distance r = 2.80mm
distance from midpoint to point A = 2.00mm
attached is the diagram of the solution, describing the position of the charge
note x = r/2, where x is the distance from midpoint of r to the particle
using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m
the electric field at point A is given as
vector <em>E </em>= 2E×cos θ( -x direction)
recall E =kq/x²
where k is the electrostatics constant = 1/4πε₀
where ε₀ is permittivity of free space
therefore using E =2{kq/x²}cosθ
∴cosθ = adjacent/hypotenuse
cosθ=1.40/2.44
E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)
E=2{4.79×10¹³}×(0.574)(-x)
E = 2×2.75 ×10¹³N/C(-x direction)
Vector <em>E= </em>5.5 ×10¹³N/C(-x direction)
Answer:
(A) Capacitance per unit length =
(B) The magnitude of charge on both conductor is C and the sign of charge on inner conductor is
and the sign on outer conductor is
Explanation:
Given :
Radius of inner part of conductor =
m
Radius of outer part of conductor =
m
The length of the capacitor =
m
(A)
Capacitance is purely geometrical property. It depends only on length, radius of conductor.
From the formula of cylindrical capacitor,
Where,
But we need capacitance per unit length so,
capacitance per unit length =
(B)
The charge on both conductors is given by,
Where, C = capacitance of cylindrical capacitor and value of F,
V
∴ C
The magnitude of charge on both conductor is same as above but the sign of charge is different.
Charge on inner conductor is and Charge on outer conductor is
.