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A 5.00 g charged insulating ball hangs on a 30.0 cm long string in a uniform horizontal electric field as shown in Figure 18.56. Given the charge on the ball is $1.00\textrm{ }\mu\textrm{C}$ , find the strength of the field.
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<b>Figure 18.56</b> A charged pith ball hanging from a string in the presence of a horizontal electric field.
Figure 18.56 A charged pith ball hanging from a string in the presence of a horizontal electric field.
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$7\times 10^{3}\textrm{ N/C}$
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OpenStax College Physics Solution, Chapter 18, Problem 60 (Problems & Exercises) (3:22)

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Video Transcript
This is College Physics Answers with Shaun Dychko. A pith ball with a 1.00 times 10 to the minus 6 coulomb charge is suspended from a 30.0 centimeter string at an angle of 8 degrees with respect to vertical and it's in the presence of this electric field which is pushing it to the right and we know that this tension force in this string has to balance both the gravity going straight down and the electric field force pointing straight to the right. So we can talk about the x-direction and that is to say that the electric field to the right has to equal the component of tension that's to the left; this is a right angle in here... in here is the angle 8 degrees because this vertical dotted line and this vertical component are parallel and these are interior opposite angles which are equal. Okay! So we can say that and substitute for each of these things: we'll write down the x-component of the tension force as the tension force multiplied by sin of Θ because this is the opposite leg of this right triangle and we can multiply the hypotenuse by sin of Θ to find the opposite leg and the electrostatic force is the magnitude of the charge multiplied by the electric field strength. And that's all we can say at that point, there are two unknowns we don't know: what the tension force is and we don't know what the electric field is and so we can't solve this equation. We need to have a second equation... you always need as many equations as you have unknowns and we have two unknowns so that means we need two equations. So then we consider the y-direction and the gravity straight down is going to be balanced by the component of tension force that's straight up— this y-component— and the y-component of tension is the tension force multiplied by cosine of Θ because this is the adjacent leg of the right triangle and we multiply the hypotenuse by cos Θ to find the adjacent leg. And then the gravity is mass times gravitational field strength, g, and now we have two equations and two unknowns and we are going to solve by dividing these two equations— that's the most convenient method. So we can take the left sides and divide them so that's qE over mg and then take the right sides and divide them that's F T sin Θ divided by F T cos Θ. So the F T's cancel and then there's a trigonometric identity which says that tangent Θ is the same as sin Θ divided by cos Θ so we make that substitution and then solve for E by multiplying both sides by mg over q. So the electric field strength then is mass times gravitation field strength times tan of this angle divided by the charge. So that's 5.00 times 10 to the minus 3 kilograms— mass of the pith ball— times 9.81 newtons per kilogram times tan of 8 degrees divided by 1.00 microcoulomb which is 1.00 times 10 to the minus 6 coulombs giving an electric field strength of 7 times 10 to the 3 newtons per coulomb and keeping only one significant figure here since our precision in the angle is only one significant figure.