In the diagram, q1 = +4.88*10^-8 C.
What is the potential difference when
you go from point A to point B?
Include the correct sign, + or - .
B
0.538 m
1.36 m
91 |
(Hint: Does V go up or down when you go
from B to A?) (Unit = V)
Answers
Answer:
ΔV = 1139.3 V = 1.139 KV (+ve sign shows V goes up)
Explanation:
The potential difference while moving from point A to Point B is given as follows:
[tex]\Delta V = V_B-V_A[/tex]
where,
ΔV = potential difference from A to B = ?
[tex]V_A[/tex] = Potential at point A = [tex]\frac{kq}{r_A}[/tex]
[tex]V_B[/tex] = Potential at point B = [tex]\frac{kq}{r_B}[/tex]
Therefore,
[tex]\Delta V = \frac{kq}{r_B}-\frac{kq}{r_A}\\\\\Delta V = kq(\frac{1}{r_B}-\frac{1}{r_A})[/tex]
where,
k = Colomb's Constant = 9 x 10⁹ N.m²/C²
q = magnitude of charge = 4.88 x 10⁻⁸ C
[tex]r_A[/tex] = distance of point A from charge = 1.36 m
[tex]r_B[/tex] = distance of point B from charge = 0.538 m
Therefore,
[tex]\Delta V = (9\ x\ 10^9\ N.m^2/C^2)(4.88\ x\ 10^{-8}\ C)(\frac{1}{0.538\ m}-\frac{1}{1.36\ m})\\\\\Delta V = (439.2 N.m^2/C)(2.59\ /m)[/tex]
ΔV = 1139.3 V = 1.139 KV (+ve sign shows V goes up)
The answer is 492.87.
Correct on Acellus
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Answers
Answer:
Direct
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Answers
We know
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[tex]\\ \sf\longmapsto W\approx 430J[/tex]
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⇒[tex]W=250\sqrt{3}[/tex]
Hence,
⇒[tex]W=250(1.73)[/tex]
[tex]W[/tex] approximately 430J.
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Answer:
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