Math

QuestionThree charges are placed at the corners of a rectangle, and observation point PP is located at the fourth, as shown. The rectangle, with a horizontal side length of h=7.11 cmh=7.11 \mathrm{~cm} and a vertical side length of v=3.91 cmv=3.91 \mathrm{~cm}, is aligned with the coordinate axes. The charges at the upper-left, upper-right, and lower-left corners are qUL=+4.27nCq_{\mathrm{UL}}=+4.27 \mathbf{n C}, qUR=7.34nCq_{\mathrm{UR}}=-7.34 \mathrm{nC}, and qLL=+5.33nCq_{\mathrm{LL}}=+5.33 \mathrm{nC}, respectively. Arrington, Parker - pjas37)epsu.edu - Part (a) \checkmark
The net electric field at point PP may be expressed in Cartesian unit-vector notation as Enet =Exi^+Eyj^\vec{E}_{\text {net }}=E_{x} \hat{i}+E_{y} \hat{j}
Enter the value, in newtons per coulomb, of the horizontal component of the net electric field. Ex=1.459×104 N/CE_{x}=1.459 \times 10^{4} \mathrm{~N} / \mathrm{C} \checkmark Correct!
Part (b) The net electric field at point PP may be expressed in Cartesian unit-vector notation as Enet =Exi^Eyj^\vec{E}_{\text {net }}=E_{x} \hat{i} \mp E_{y} \hat{j}
Enter the value, in newtons per coulomb, of the vertical component of the net electric field. Ey=E_{y}= \square N/C\mathrm{N} / \mathrm{C} \square ( \square 7 ( 7 8 9 \square sinθcosθtanθcotanθasinθacosθatanθacotanθsinhθcoshθtanhθcotanhθ\begin{array}{|l|l|l|} \sin \theta & \cos \theta & \tan \theta \\ \hline \operatorname{cotan} \theta & a \sin \theta & a \cos \theta \\ \hline \operatorname{atan} \theta & \operatorname{acotan} \theta & \sinh \theta \\ \hline \cosh \theta & \tanh \theta & \operatorname{cotanh} \theta \\ \hline \end{array} E 13 (1) 4 5 6 0 \square Degrees Radians 1 \square 1 2 3 \square \square \square \square END ++++ 0-0 ((. CLEAR Submit \square PACKspace \square \square (V) Feedback: I give up! 2 Free Submission(s) Remaining Hints: 2%\underline{2 \%} deduction per hint. Hints remaining: 3\underline{3} Feedback: 2%2 \% deduction per feedback. Part (c) What is the magnitude, in newtons per coulomb, of the net electric field at point PP.

Studdy Solution
Calculate the magnitude of the net electric field at point PP. Enet=Ex2+Ey2 E_{\text{net}} = \sqrt{E_{x}^2 + E_{y}^2} Ex=1.459×104N/C E_{x} = 1.459 \times 10^4 \, \text{N/C} Enet=(1.459×104)2+(4.01×104)24.26×104N/C E_{\text{net}} = \sqrt{(1.459 \times 10^4)^2 + (-4.01 \times 10^4)^2} \approx 4.26 \times 10^4 \, \text{N/C}
The vertical component of the net electric field is: Ey=4.01×104N/C E_{y} = -4.01 \times 10^4 \, \text{N/C}
The magnitude of the net electric field at point PP is: Enet=4.26×104N/C E_{\text{net}} = 4.26 \times 10^4 \, \text{N/C}

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