Problem 8: Four point charges of equal magnitude Q = 35 nC are placed on the corners of arectangle of sides D1 = 22 cm and D, = 9 cm. The charges on the left side of the rectangle arepositive while the charges on the right side of the rectangle are negative. Refer to the figure.+QPart (e) Calculate the value of the vertical component of the net force, in newtons.F, =D2+QDIPart (f) Calculate the magnitude of the net force, in newtons, on the charge located at the lower left corner of the rectangle.F =Part (g) Calculate the angle, in degrees between -180° and +180°, that the net force makes, measured from the positive horizontal direction.

Answered: Problem S: Four point charges of equal…

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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Problem 8: Four point charges of equal magnitude Q = 35 nC are placed on the corners of a
rectangle of sides D1 = 22 cm and D, = 9 cm. The charges on the left side of the rectangle are
positive while the charges on the right side of the rectangle are negative. Refer to the figure.
+Q
Part (e) Calculate the value of the vertical component of the net force, in newtons.
F, =
D2
+Q
DI
Part (f) Calculate the magnitude of the net force, in newtons, on the charge located at the lower left corner of the rectangle.
F =
Part (g) Calculate the angle, in degrees between -180° and +180°, that the net force makes, measured from the positive horizontal direction.

Expert Solution

Step 1

a)To find the net force, Force between Pairs at the sides and the diagonal pairs must be calculated.

Net force,

$F = F_{1} + F_{2} + F_{3} + F_{4} + F_{5} + F_{6} = - F_{1} J + F_{2} i + F_{3} j - F_{4} i + F_{5} \cos 45^{0} i + F_{5} \sin 45^{0} j - F_{6} \cos 45^{0} i + F_{6} \sin 45^{0} j \frac{}{} = {\frac{- k {(35 \times 10^{- 9})}^{2}}{9^{2}} + \frac{k (35 \times 10^{- 9})}{9^{2}} + 2 (\frac{k {(35 \times 10^{- 9})}^{2}}{{(\sqrt{22^{2^{}} + 9^{2}})}^{2}} \times \sin 45)} j + {\frac{k {(35 \times 10^{- 9})}^{2}}{22^{2}} - \frac{k {(35 \times 10^{- 9})}^{2}}{22^{2}} + \frac{k {(35 \times 10^{- 9})}^{2}}{{(\sqrt{22^{2} + 9^{2}})}^{2}} \cos 45 - \frac{k {(35 \times 10^{- 9})}^{2}}{{(\sqrt{22^{2^{}} + 9^{2}})}^{2}} \cos 45} i$

The horizontal component of net force is

$= 2 \times \frac{k {(35 \times 10^{- 9})}^{29}}{{(\sqrt{22^{2} + 9^{2}})}^{2}} \times \sin 45^{0} i = 27.6 \times 10^{- 9} i N$

Magnitude is $27.6 \times 10^{- 9} N$

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