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Answer :
Final answer:
To find the electric potential between two coaxial rings of charge, utilize the formula for electric potential due to a ring of charge and sum up potentials from infinitesimal charges within the rings.
Explanation:
Electric potential at a point on the axis midway between two coaxial rings of charge can be calculated using the formula for electric potential due to a ring of charge.
Given the radii, separation distance, and charge, you can determine the potential at the midway point using the provided Coulomb constant.
By summing up the potentials from infinitesimal charges within the rings, you can find the electric potential at the point of interest.
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The electric potential at the midpoint on the common axis is approximately 1.66 x [tex]10^5[/tex] volts.
The electric potential at the midpoint on the common axis of two coaxial rings can be calculated by considering the individual contributions of each ring and summing them up. Here's how:
Given:
Radius of each ring (a) = 15.2 cm = 0.152 m
Distance between the rings (d) = 36.3 cm = 0.363 m
Charge on each ring (Q1 & Q2) = 2.18 C
Coulomb constant (k) = 8.98755 x [tex]10^9[/tex] N.m²/C²
Steps:
Calculate the distance from the midpoint to each ring:
This distance is half the separation between the rings.
Distance (r) = d/2 = 0.363 m / 2 = 0.1815 m
Apply the formula for electric potential due to a charged ring:
V = k × Q / (2 × π × ε₀ × r)
Where:
V is the electric potential
k is the Coulomb constant
Q is the charge on the ring
ε₀ is the permittivity of free space (8.854 x [tex]10^-12[/tex] F/m)
r is the distance from the ring to the point
Calculate the potential due to each ring:
V1 = k × Q1 / (2 × π × ε₀ × r) = 8.98755 x [tex]10^9[/tex] N.m²/C² × 2.18 C / (2 × π × 8.854 x 10^-12 F/m × 0.1815 m)
V2 (similarly for the second ring) = 8.98755 x [tex]10^9[/tex] N.m²/C² × 2.18 C / (2 × π × 8.854 x 10^-12 F/m × 0.1815 m)
Sum the potentials due to both rings:
V_total = V1 + V2
Calculation:
V1 ≈ V2 ≈ 8.284 x 10^4 V (rounded to 4 significant figures)
V_total ≈ V1 + V2 ≈ 8.284 x [tex]10^4[/tex] V + 8.284 x [tex]10^4[/tex] V ≈ 1.66 x [tex]10^5 V[/tex]
Therefore, the electric potential at the midpoint on the common axis is approximately 1.66 x [tex]10^5[/tex] volts.
