A line of positive charge is formed into a semicircle of radius R = 60.0 cm as shown in Figure P23.41.
A line of positive charge is formed into a semicircle of radius R = 60.0 cm as shown in Figure P23.41. The charge per unit length along the semicircle is given by the expression λ = λ0cos θ. The total charge on the semicircle is 12.0 μC. Calculate the total force on a charge of 3.00 μC placed at the center of curvature P.
Homework Answers
Electric field :
Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge.
It is given by the formula :
Force :
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A line of positive charge is formed into a semicircle of radius R = 60.0 cm as shown in Figure P23.41.
A line of postive charge formed into a semicircle of radius R = 50.0 cm, as shown in the figure...
A line of postive charge formed into a semicircle of
radius R = 50.0 cm, as shown in the figure
below. The charge per unit length along the semicircle is described
by the expression λ = λ0 cos(θ). The total
charge on the semicircle is 12.0 µC. Calculate the total
force on a 3.00 µC. placed at the centet of curvature.
A line of positive charge is formed into a semicircle of radius R 80.0 cm, as...
A line of positive charge is formed into a semicircle of radius R 80.0 cm, as shown in the figure below. The charge per unit length along the semicircle is given by the expression λ-, cos@). The total charge on the semicircle is 13.0 μC. Calculate the total force on a charge of 2.00 μC placed at the center of curvature magnitude direction Select
A line of positive charge is formed into a semicircle of radius R 80.0 cm, as...
A line of positive charge is formed into a semicircle of radius R 80.0 cm, as shown in the figure below. The charge per unit length along the semicircle is given by the expression λ,cos(8). The total charge on the semicircle is 13.0 μC. Calculate the total force on a charge of 2.00 μC placed at the center of curvature. magnitude Poor response differs from the correct answer by more than 100%. direction upward
A line of positive charge is formed into a semicircle of radius R 57.4cm, as shown...
A line of positive charge is formed into a semicircle of radius R 57.4cm, as shown in the figure below. The charge per unit length along the semicircle is described by the expression charge on the semicircle is 13.5uC. Calculate the value of the constant A Submit Answer Tries o/10 Calculate the magnitude of the total force on a charge of 2.94uC placed at the center of curvature. (8) λ0cose. The total Submit Answer Tries 0/10
A line of positive charge is formed into a semicircle of radius R = 45.0 cm,...
A line of positive charge is formed into a semicircle of radius R = 45.0 cm, as shown in the figure below. The charge per unit length along the semicircle is described by the expression ? = ?0 cos(?). The total charge on the semicircle is 16.0
Three point charges, q1 = -4.19nC, q2 = 5.31nC and q3 = 2.70nC are aligned along the x axis as shown in the figure below. Assume that L1 = 0.545m and L2 = 0.755m.
Three point charges, q1 = -4.19nC, q2 = 5.31nC and q3 = 2.70nC are aligned along the x axis as shown in the figure below. Assume that L1 = 0.545m and L2 = 0.755m. Calculate the electric field at the position (2.13m, 0). Calculate the magnitude of the electric field at the position (0, 2.13m). Calculate the angle of the electric field with respect to the positive x-axis at this position (0, 2.13). A line of positive charge is formed into a semicircle of...
I don't understand why my solution did not work, I tried writing a function that gives...
I don't understand why my
solution did not work, I tried writing a function that gives the
charge density with the variable "L", where L is the length along
the circumference of the circle. Then I attempted to integrate from
the lowest possible L, 0, to the highest, piR, but my answer came
out as 0 (wrong). Could someone please explain to me the flaw in my
method?
A line of positive charge is formed into a semicircle of radius...
In the figure a thin glass rod forms a semicircle of radius r = 2.41 cm....
In the
figure a thin glass rod forms a semicircle of radius
r
= 2.41 cm.
Charge is uniformly distributed along the rod, with
+q
= 1.66 pC
in the upper half and -q
= -1.66 pC
in the lower half. What is the magnitude of the electric field
at P,
the center of the semicircle?
In the figure a thin glass rod forms a semicircle of radius r = 2.41 cm. Charge is uniformly distributed along the rod, with...
In the figure a thin glass rod forms a semicircle of radius r = 2.09 cm....
In the figure a thin glass rod forms a semicircle of radius
r = 2.09 cm. Charge is uniformly distributed along the
rod, with +q = 1.05 pC in the upper half and -q =
-1.05 pC in the lower half. What is the magnitude of the electric
field at P, the center of the semicircle?
In the figure a thin glass rod forms a semicircle of radius r = 2.09 cm. Charge is uniformly distributed along the rod, with...
Problem 21.50 A thin glass rod is a semicircle of radius R, see the figure Tap...
Problem 21.50 A thin glass rod is a semicircle of radius R, see the figure Tap image to zoom А charge is nonuniformly distributed along the rod with a linear charge density given by λ-Aa sin θ , where λο is a positive constant. Point P is at the center of the semicircle. Part A Find the electric field E (magnitude and direction) at point P [Hint Remember sin(-0)--sin θ , so the two halves of the rod are oppostely...
A line of postive charge formed into a semicircle of
radius R = 50.0 cm, as shown in the figure
below. The charge per unit length along the semicircle is described
by the expression λ = λ0 cos(θ). The total
charge on the semicircle is 12.0 µC. Calculate the total
force on a 3.00 µC. placed at the centet of curvature.
A line of positive charge is formed into a semicircle of radius R 80.0 cm, as shown in the figure below. The charge per unit length along the semicircle is given by the expression λ-, cos@). The total charge on the semicircle is 13.0 μC. Calculate the total force on a charge of 2.00 μC placed at the center of curvature magnitude direction Select
A line of positive charge is formed into a semicircle of radius R 80.0 cm, as shown in the figure below. The charge per unit length along the semicircle is given by the expression λ,cos(8). The total charge on the semicircle is 13.0 μC. Calculate the total force on a charge of 2.00 μC placed at the center of curvature. magnitude Poor response differs from the correct answer by more than 100%. direction upward
A line of positive charge is formed into a semicircle of radius R 57.4cm, as shown in the figure below. The charge per unit length along the semicircle is described by the expression charge on the semicircle is 13.5uC. Calculate the value of the constant A Submit Answer Tries o/10 Calculate the magnitude of the total force on a charge of 2.94uC placed at the center of curvature. (8) λ0cose. The total Submit Answer Tries 0/10
I don't understand why my
solution did not work, I tried writing a function that gives the
charge density with the variable "L", where L is the length along
the circumference of the circle. Then I attempted to integrate from
the lowest possible L, 0, to the highest, piR, but my answer came
out as 0 (wrong). Could someone please explain to me the flaw in my
method?
A line of positive charge is formed into a semicircle of radius...
In the
figure a thin glass rod forms a semicircle of radius
r
= 2.41 cm.
Charge is uniformly distributed along the rod, with
+q
= 1.66 pC
in the upper half and -q
= -1.66 pC
in the lower half. What is the magnitude of the electric field
at P,
the center of the semicircle?
In the figure a thin glass rod forms a semicircle of radius r = 2.41 cm. Charge is uniformly distributed along the rod, with...
In the figure a thin glass rod forms a semicircle of radius
r = 2.09 cm. Charge is uniformly distributed along the
rod, with +q = 1.05 pC in the upper half and -q =
-1.05 pC in the lower half. What is the magnitude of the electric
field at P, the center of the semicircle?
In the figure a thin glass rod forms a semicircle of radius r = 2.09 cm. Charge is uniformly distributed along the rod, with...
Problem 21.50 A thin glass rod is a semicircle of radius R, see the figure Tap image to zoom А charge is nonuniformly distributed along the rod with a linear charge density given by λ-Aa sin θ , where λο is a positive constant. Point P is at the center of the semicircle. Part A Find the electric field E (magnitude and direction) at point P [Hint Remember sin(-0)--sin θ , so the two halves of the rod are oppostely...
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