The total charge on the semi circle is given to
be .
That is
For one quadrant (ie from y axis to x axis), since the charge distribution is dependent on theta, we can find the charge in one half of this semi cirlce and then multiply it with 2 to get the total charge.
so
Now, the x component of the force experienced by the by
the component of the semi circle to the left of y axis and to the
right of y axis will be the same, but in opposite direction.
Therefore, they will cancel out each other. So only the y component
of the force will remain( this is because
but
, thus cancels each other out).
The y component force is given by the equation
here we substituted the value for dl as we took earlier.
solving, we get
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 = 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.
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 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, 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. 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...
6. (20 points) A positive charge of 2.00 uC is distributed uniformly around a semicircle of radius 2.00cm as shown. (a) Find the line charge density of the semicircle. b) Find the electric field (direction & magnitude) at the center of curvature O 0-
An infinite line of positive charge lies along the y axis, with charge density λ = 2.30 μC/m. A dipole is placed with its center along the x axis at x = 28.0 cm. The dipole consists of two charges ±10.0 μC separated by 2.00 cm. The axis of the dipole makes an angle of 45.0° with the x axis, and the positive charge is farther from the line of charge than the negative charge. Find the net force exerted...
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 4.00 cm. charge is uniformly distributed along the rod, with +9 = 6.00 pC in the upper half and-q =-6.00 pC in the lower half. (a) What is the magnitude of the electric field at P, the center of the semicircle? N/C (b) What is its direction? counterclockwise from the +x-axis