Homework Answers
force acting on negative
charge is opposite to electric field so force on electron is along
positive y axis i.e. up
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A thin glass nod is a of radius R see the fgureFigure 1) A where λ's...
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...
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...
In the figure a thin glass rod forms a semicircle of radius r = 2.35 cm....
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....
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?
Question6 In the figure a thin glass rod forms a semicircle of radius r- 2.37 cm....
Question6 In the figure a thin glass rod forms a semicircle of radius r- 2.37 cm. Charge is uniformly distributed along the rod, with +q1.17 pC in the upper half and-q =-1.17 pC in the lower half. What is the magnitude of the electric field at P, the center of the semicircle? Number Units
In the figure, a thin glass rod forms a semicircle of radius r = 2.50 cm....
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....
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 Fig. 22-44, a thin glass rod forms a semicircle of radius r = 3.87 cm....
In Fig. 22-44, a thin glass rod forms a semicircle of radius r
= 3.87 cm. Charge is uniformly distributed along the rod, with +q =
3.52 pC in the upper half and -q = -3.52 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 4.00 cm. charge...
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
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
= 2.41 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...
Question6 In the figure a thin glass rod forms a semicircle of radius r- 2.37 cm. Charge is uniformly distributed along the rod, with +q1.17 pC in the upper half and-q =-1.17 pC in the lower half. What is the magnitude of the electric field at P, the center of the semicircle? Number Units
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 Fig. 22-44, a thin glass rod forms a semicircle of radius r
= 3.87 cm. Charge is uniformly distributed along the rod, with +q =
3.52 pC in the upper half and -q = -3.52 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 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
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