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cluponka [151]
3 years ago
12

9. From this lab, we learn that the electric field and electric potential depend on both, the magnitude of the source charge (q)

, and the distance from the source charge (r). If we were to increase the magnitude of our source charge from 1 nC to 5 nC, then the magnitudes of the electric field and electric potential would be ____.(you can test this on the animation by dragging five 1 nC charges on top of each other and measuring E and V at a distance of 1 m)
Physics
1 answer:
VARVARA [1.3K]3 years ago
5 0

Answer:

the electric field and the electric potential increase 5 times

Explanation:

The electric field created by a point charge is

         E = k q / r²

in this case the charge changes from q₁ = 1 10⁺⁰ C to q₂ = 5 10⁻⁹ C

with the electric field is proportional to the charge

         E₅ = 5 E₁

the electrical power for a point charge is

         V = k q / r

as the electric power is proportional to the charge

         V₅ = 5 V₁

consequently both the electric field and the electric potential increase 5 times

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The Earth orbits around the sun. Which scenario would cause the
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2. A 15 kg mass fastened to the end of a steel wire of un-
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Explanation:

Elongation of the wire is:

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where F is the force,

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E is Young's modulus,

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ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)

ΔL = T (1.25×10⁻⁶ m/N)

T = (80,000 N/m) ΔL

Draw a free body diagram of the mass at the bottom of the circle.  There are two forces: tension force T pulling up and weight force mg pulling down.

Sum of forces in the centripetal direction:

∑F = ma

T − mg = mv²/r

T − mg = mω²r

T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)

T − 147 N = (2368.7 N/m) (0.5 m + ΔL)

Substitute:

(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)

(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL

(797631.3 N/m) ΔL = 1331.35 N

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