You have a suspect on the run in a dry, wooded area; they’ve ditched the gun but are hiding from you. how long do you have to fi
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Answer:
the electric potential difference between the point at the center of the ring and a point on its axis ΔV is
Explanation:
Given the data in the question;
electric potential at the center of the ring V₀ = kQ / R
electric potential on the axis point Vr = kQ / √( R² + x² )
at a distance 6R from the center,
point at x = 6R
so distance circumference r = √( R² + (6R)² )
so
electric potential on the axis point Vr = kQ / √( R² + (6R)² )
Vr = kQ / R√37
Now
ΔV = V₀ - Vr
we substitute
ΔV = ( kQ / R) - ( kQ / R√37 )
ΔV = kQ/R( 1 - 1/√37 )
ΔV = kQ/R( 1 - 0.164398987 )
ΔV = kQ/R( 0.8356 )
ΔV = { where k =
}
Therefore, the electric potential difference between the point at the center of the ring and a point on its axis ΔV is