In elastic collisions,total kinetic energy and momentum are both conserved. On a frictionless track,cart 1 is moving with a cons
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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)
