Solution,
Electric field, Ey = 2 |E| cos45
Ey = 2[(1/4 pi e0) x 2 sqrt(2) x q/ pi r^2] cos45
Ey = 4[(9 x 10^9) x (5.98 x 10^-12)/ pi x (0.0275)^2
Ey = 90.61 N/C
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Chapter 22, Problem 026 In the figure a thin glass rod forms a semidircle of radlus...
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
= 2.41 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 with tq 5.46 pC in the upper half and-q5.46 pC in the lower half. What is the magnitude of the electric field at P, the center of the semicircle? semicircle of radius r 2.51 cm. Charge is uniformly distributed along the rod
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?
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?
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
Send to Gradebook K Prev Next > Question 6 In the figure a thin glass rod forms a semicircle of radius r- 1.90 cm. Charge is uniformly distributed along the rod, with +q = 4.24 pC in the upper half and-q =-4.24 pC in the lower half. What is the magnitude of the electric field at P, the center of the semicircle? -9 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