Temperature, gravity, atmosphere and water
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Answer:
0.0170m or 1.70cm
Explanation:
L=(μ₀*N^2*A)/(l)
N=x/2*pi*r so r=x/2*pi*N
A=pi*r^2=(pi*x^2)/(4*pi^2*N^2)
L=(μ₀*N^2*A)/(l)=(μ₀*N^2*pi*x^2)/(4*pi^2*N^2*l)=(μ₀*x^2)/(4*pi*l)
l=(μ₀*x^2)/4*pi*L)=[(4*10^-7)*(1.15m)/4*pi*2.47*10^-6H)=0.0170m or 1.70cm
Hope this helps. Any questions please feel free to ask. Thanks!
Answer:
q = 4.5 nC
Explanation:
given,
electric field of small charged object, E = 180000 N/C
distance between them, r = 1.5 cm = 0.015 m
using equation of electric field

k = 9 x 10⁹ N.m²/C²
q is the charge of the object

now,

q = 4.5 x 10⁻⁹ C
q = 4.5 nC
the charge on the object is equal to 4.5 nC