A small 12.0 g plastic ball is tied to a very light 25.7 cm string that is attached to the vertical wall of a room. (See the figure ( Figure 1).) A uniform horizontal electric field exists in this room. When the ball has been given an excess charge of -1.20 μC, you observe that it remains suspended, with the string making an angle of 17.4° with the wall. Figure 1 of 1 ▾ Part A Find the magnitude of the electric field in the room. Express your answer in newtons per coulomb. ID ΑΣΦ E= Submit Part B Request Answer Find the direction of the electric field in the room. to the left g ? N/C
A small 12.0 g plastic ball is tied to a very light 25.7 cm string that is attached to the vertical wall of a room. (See the figure ( Figure 1).) A uniform horizontal electric field exists in this room. When the ball has been given an excess charge of -1.20 μC, you observe that it remains suspended, with the string making an angle of 17.4° with the wall. Figure 1 of 1 ▾ Part A Find the magnitude of the electric field in the room. Express your answer in newtons per coulomb. ID ΑΣΦ E= Submit Part B Request Answer Find the direction of the electric field in the room. to the left g ? N/C
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:A small 12.0 g plastic ball is tied to a very light 25.7 cm string
that is attached to the vertical wall of a room. (See the figure (
Figure 1).) A uniform horizontal electric field exists in this
room. When the ball has been given an excess charge of
-1.20 μC, you observe that it remains suspended, with the
string making an angle of 17.4° with the wall.
Figure
17.4°
1 of 1
Part A
Find the magnitude of the electric field in the room.
Express your answer in newtons per coulomb.
E =
Submit
Part B
Find the direction of the electric field in the room.
to the left
Request Answer
to the right
Submit
ΑΣΦ
Provide Feedback
Request Answer
?
N/C
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Expert Solution

Step 1 GIVEN
Mass, m = 12 g = 12×10-3 kg = 0.012 kg
Length, l = 25.7 cm
Charge, q = -1.20 µC = -1.20×10-6 C
Angle, θ = 17.4°
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