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Oaklandnachos9762
07/22/2020
• Physics • High School
answered • expert verified
A +35 µC point charge is placed 46 cm from an identical +35 µC charge.
How much work would be required to move a +0.50 µC test charge from a
point midway between them to a point 12 cm closer to either of the
charges?
2
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throwdolbeau
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• 4.6K answers • 31M people helped
Work done will be:
.
Given:
q = +35 µC
q' = +0.50 µC
r₁ = 46 cm
r₂ = 46 cm/2 = 23 cm = 0.23 m
The electric potential at this point due to the two charges q is thus
When the charge q' is moved 12 cm closer to either of the two charges, its
distance from each charge is now;
r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm = 0.35 m and
r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.
Thus, the new electric potential at this point is
Now, the work done in moving the charge q' to the point 12 cm from either
charge is:
Thus, the work done will be:
.
Find out more information on "Work done" here:
brainly.com/question/25573309
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Answer
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oladayoademosu
Virtuoso
• 2.3K answers • 3.9M people helped
Answer:
512.5 mJ
Explanation:
Let the two identical charges be q = +35 µC and distance between them be
r₁ = 46 cm. A charge q' = +0.50 µC located mid-point between them is at r₂ =
46 cm/2 = 23 cm = 0.23 m.
The electric potential at this point due to the two charges q is thus
V = kq/r₂ + kq/r₂
= 2kq/r₂
= 2 × 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C/0.23 m
= 630/0.23 × 10³ V
= 2739.13 × 10³ V
= 2.739 MV
When the charge q' is moved 12 cm closer to either of the two charges, its
distance from each charge is now r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm =
0.35 m and r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.
So, the new electric potential at this point is
V' = kq/r₃ + kq/r₄
= kq(1/r₃ + 1/r₄)
= 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C(1/0.35 m + 1/0.11 m)
= 315 × 10³(2.857 + 9.091) V
= 315 × 10³ (11.948) V
= 3763.62 × 10³ V
= 3.764 MV
Now, the work done in moving the charge q' to the point 12 cm from either
charge is
W = q'(V' - V)
= 0.5 × 10⁻⁶ C(3.764 MV - 2.739 MV)
= 0.5 × 10⁻⁶ C(1.025 × 10⁶) V
= 0.5125 J
= 512.5 mJ
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