y(0, +4.0 m)Q(-3.0 m, 0)(3.0 m. 0)(0,-8.0 m)BEach Q shown is 60 µC. What work is required of some agent to move a charge q = 5 µC frompoint A to point B without accelerating q?J aa+2QThree colinear point charges are positioned as shown. If Q = 21 nC, a = 3 cm, and b = 5 cm,what is the electric potential at point P, with the convention that the potential is zero at infinity?%3DkV

Answered: Each Q shown is 60 µC. What work is…

Each Q shown is 60 µC. What work is required of some agent to move a charge q = 5 µC fror point A to point B without accelerating q? %3D

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Question

y
(0, +4.0 m)
Q
(-3.0 m, 0)
(3.0 m. 0)
(0,-8.0 m)
B
Each Q shown is 60 µC. What work is required of some agent to move a charge q = 5 µC from
point A to point B without accelerating q?
J

a
a
+2Q
Three colinear point charges are positioned as shown. If Q = 21 nC, a = 3 cm, and b = 5 cm,
what is the electric potential at point P, with the convention that the potential is zero at infinity?
%3D
kV

Expert Solution

Step 1

Since we answer up to one question, we will answer the first question only. Please resubmit the question and specify the other question you would like to get answered.

Given: The value of the two charge Q is $60 μC or 60 \times 10^{- 6} C$ .

To determine: The work required to move a charge q of $5 μC or 5 \times 10^{- 6} C$ from point A to B.

The figure shows the distance between the charges to point A and B

The distance between A and C is calculated using Pythagoras rule,

$AC = \sqrt{{(3 m)}^{2} + {(4 m)}^{2}} AC = 5 m$

Similarly value of AD is also 5 m.

The distance C and B is

$CB = \sqrt{{(3 m)}^{2} + {(6 m)}^{2}} CB = 6.71 m$

Similarly the distance between DB is 6.71 m.

The potential is given as

$V = \frac{kQ}{r}$

where V is the potential, k is the constant, value of k is $9 \times 10^{9} N . m^{2} / C^{2}$ and r is the distance.

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