A thin glass rod is a semi-circle of radius R. a charge is non-uniformly distributes along...
A thin glass rod forms a semicircle of radius r= 5.00 cm. Charge is uniformly distributed along the rod, with +q = 4.50 pC in the upper half and -q = -4.50 pC in the lower half. What are the: (a) Magnitude and: (b) Direction (relative to the positive direction of the x axis) of the electric field E 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 +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...
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
4. (5 points) A thin glass rod is bent into a semicircle of radius r. A charge +q is uniformly distributed along the upper half and a charge -is uniformly distributed on the lower half. Find the electric field E at p (0,0,0), which is center of the semi-cirde.
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...
In the figure, a thin glass rod forms a semicircle of radius r = 2.50 cm. Charge is uniformly distributed along the rod, with +q = 5.50 pC in the upper half and −q = −5.50 pC in the lower half. (a) What is the magnitude of the electric field at P, the center of the semicircle? (b) What is its direction? ???° counterclockwise from the +x-axis +q -9
In the figure, a thin glass rod forms a semicircle of radius r = 2.50 cm. Charge is uniformly distributed along the rod, with +q = 5.50 pC in the upper half and −q = −5.50 pC in the lower half. (a) What is the magnitude of the electric field at P, the center of the semicircle? (b) What is its direction? ° counterclockwise from the +x-axis +q -9
In the figure a thin glass rod forms a semicircle of radius r = 2.35 cm. Charge is uniformly distributed along the rod, with +q = 3.49 pC in the upper half and -q = -3.49 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 = 5.75 cm. Charge is uniformly distributed along the rod, with +q = 4.03 pC in the upper half and -q = - 4.03 pC in the lower half. What is the magnitude of the electric field at P, the center of the semicircle?