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Question

Positive charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positive x-axis at x=a+r, a distance r to the right of the end of Q. (Figure 1)

Part A
Calculate the x-component of the electric field produced by the charge distribution Q
at points on the positive x-axis where x>a.
Express your answer in terms of the variables
Q, a, x, r, and appropriate constants.


Part B
Calculate the y-component of the electric field produced by the charge distribution Q at points on the positive x-axis where x>a.
Express your answer in terms of the variables
Q, a, x, r, and appropriate constants.

Part C
Calculate the magnitude of the force that the charge distribution Q exerts on q.
Express your answer in terms of the variables Q.

Part D
In what direction the charge distribution Q exerts force on q.
In what direction the charge distribution exerts force on .to the left to the right

 

 

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x
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check_circleAnswer
Step 1

(A)

The equation for the electric field due to the charge distribution is given by

kQ r" cos0i +r' sin0j)
Here, E is the electric field, k is the Coulomb constant (1/47E0), ' is the position vector from the
charge distribution to the point where the filed has to be calculated, r' is the magnitude of F and 0 is
the angle the electric filed makes with the positive x-axis
Since the electric field is along x-axis, the angle the filed makes with the x-axis will be zero deg
Then according to the above equation, the y-component will be zero since sin 0 0.The electric field
will be due to the charge distribution on the points on x-axis will be directed along the x-axis. It will
have only x component and the y component will be zero.
e.
The center of the charge distribution is at x a/2. The value of r' for the points on the positive x-axis
is equal to x-a/2 where x is the distance from the origin to the point. The equation for the x
component of the electric field due to the charge distribution at points on the x-axis is given by
Q
E k-
(x-(a/2)
Here, E is the x-component of the electric field due to the charge distribution.
(1)
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kQ r" cos0i +r' sin0j) Here, E is the electric field, k is the Coulomb constant (1/47E0), ' is the position vector from the charge distribution to the point where the filed has to be calculated, r' is the magnitude of F and 0 is the angle the electric filed makes with the positive x-axis Since the electric field is along x-axis, the angle the filed makes with the x-axis will be zero deg Then according to the above equation, the y-component will be zero since sin 0 0.The electric field will be due to the charge distribution on the points on x-axis will be directed along the x-axis. It will have only x component and the y component will be zero. e. The center of the charge distribution is at x a/2. The value of r' for the points on the positive x-axis is equal to x-a/2 where x is the distance from the origin to the point. The equation for the x component of the electric field due to the charge distribution at points on the x-axis is given by Q E k- (x-(a/2) Here, E is the x-component of the electric field due to the charge distribution. (1)

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Step 2

B)The equation for the y-component of the electric field due to the charge distribution at points on the x-axis is given by

E0
Here, E is the x-component of the electric field due to the charge distribution
(C)
The equation for the magnitude of the force exerted by the charge distribution Q on q is given by
F qE
Here, F is the magnitude of the force exerted by the charge distribution Q on q and E is the magnitude
of the electric field exerted by Q,
E has only x-component so that
F qE
Put equation (1) in the above equation
F = k
(x-(a/2))
The charge q is at a distance x = a r from the origin
Substitute ar for x in the above equation.
F =k.
((a+r)-(a/2))
= k
(2)
(a/2)+ r)
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E0 Here, E is the x-component of the electric field due to the charge distribution (C) The equation for the magnitude of the force exerted by the charge distribution Q on q is given by F qE Here, F is the magnitude of the force exerted by the charge distribution Q on q and E is the magnitude of the electric field exerted by Q, E has only x-component so that F qE Put equation (1) in the above equation F = k (x-(a/2)) The charge q is at a distance x = a r from the origin Substitute ar for x in the above equation. F =k. ((a+r)-(a/2)) = k (2) (a/2)+ r)

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Step 3

(D)Since the charge distribution Q and the charge q are positive, the force between them will be repulsive. The...

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Electric Charges and Fields

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