Suppose that the charges in Sample Problems 1.4 item number 3 are situated as shown, find the resultant electric force on the charge at A. The diagrams are attached herewith. Suppose that the charges in Sample Problems 1.4 item number 3 aresituated as shown. Find the resultant electric force on the charge at A.3.+q1.00 mAВ1.00 mDb.+q Four point charges (two with q=2.50×10-6C and two withq=-2.50×10-6 C) are situated at the corners of a square of side1.00 m as shown. Find the resultant force that the charge at A willexperience due to the charges at the other corners of the square.+93.+qGiven: 9,=9B=+2.50×10¬6 Cr=1.0 mIc=9p=-2.50×10-6CSolution:Eq. (1.1) is used to solve for the magnitude of the electric force exerted by each of thecharges at the other corners of the square on the charge at A.Fs on A = k-|(+2.50×10¬°C)(+2.50×10¯“C)|= 19x10° N-m²/C"|(1.00 m)²= 0.05625 N20.06 NFeom A = k|(+2.50×10“CX-2.50×10°C=19x10° N m²/C²|(1.00 m)?= 0.05625 N20.06 NSolve first for the distance between D and A. Using the Pythagorean theorem,'D on A =("B on AV(1.00 m)’ +(1.00 m)² = 1.414 m~1.41 m.Therefore,Fo on A = k-=[9x10° N m²/C²| (+2.50×10°C)X-2.50×10°C||(1.414 m)²= 0.02813 N20.03 N.FB on A is repulsive. Since the charges are both positive, the chargeat B will push the charge at A away from it. Therefore, FB on A is directedto the left. Fcon A is attractive since the charges are opposite. The chargeat C will attract the charge at A toward it. Therefore, Fc on A is directeddown. Fpon A is attractive since the charges are opposite. The charge at Dwill attract the charge at A toward it. Therefore, F on A is directed downat 45° with the +x-axis, with q, placed at the origin of the x-y plane asshown in the diagram on the right. All other forces are also shown inthis diagram.FoomaFoonaFcon AThe table below presents the respective horizontal and vertical components of these forcesand the resultant electric force.ForceHorizontal ComponentVertical ComponentFoonA = 0.05625 N–0.05625 NFconA=0.05625 N–0.05625 NFoona = 0.02813 N+0.02813 N (cos 45°) =+0.0199-0.02813 N (sin 45°)=-0.0199 NResultant electric force FEF,=-0.03635 NEF,=-0.07615 NSolving for the magnitude of the resultant electric force,F =V(-0.03635 N)' +(-0.07615 N)²= 0.0844 N.Solving for the direction,|EF,0 = tanEFTA--0.07615 N64,5°= tan|-0.03635 N= 64.5°.Thus, the resultant force is 0.0844 N directed downat 64.5° with the negative x-axis as shown.

Question

Suppose that the charges in Sample Problems 1.4 item number 3 are situated as shown, find the resultant electric force on the charge at A. The diagrams are attached herewith. Suppose that the charges in Sample Problems 1.4 item number 3 aresituated as shown. Find the resultant electric force on the charge at A.3.+q1.00 mAВ1.00 mDb.+q Four point charges (two with q=2.50×10-6C and two withq=-2.50×10-6 C) are situated at the corners of a square of side1.00 m as shown. Find the resultant force that the charge at A willexperience due to the charges at the other corners of the square.+93.+qGiven: 9,=9B=+2.50×10¬6 Cr=1.0 mIc=9p=-2.50×10-6CSolution:Eq. (1.1) is used to solve for the magnitude of the electric force exerted by each of thecharges at the other corners of the square on the charge at A.Fs on A = k-|(+2.50×10¬°C)(+2.50×10¯“C)|= 19x10° N-m²/C&#34;|(1.00 m)²= 0.05625 N20.06 NFeom A = k|(+2.50×10“CX-2.50×10°C=19x10° N m²/C²|(1.00 m)?= 0.05625 N20.06 NSolve first for the distance between D and A. Using the Pythagorean theorem,'D on A =(&#34;B on AV(1.00 m)’ +(1.00 m)² = 1.414 m~1.41 m.Therefore,Fo on A = k-=[9x10° N m²/C²| (+2.50×10°C)X-2.50×10°C||(1.414 m)²= 0.02813 N20.03 N.FB on A is repulsive. Since the charges are both positive, the chargeat B will push the charge at A away from it. Therefore, FB on A is directedto the left. Fcon A is attractive since the charges are opposite. The chargeat C will attract the charge at A toward it. Therefore, Fc on A is directeddown. Fpon A is attractive since the charges are opposite. The charge at Dwill attract the charge at A toward it. Therefore, F on A is directed downat 45° with the +x-axis, with q, placed at the origin of the x-y plane asshown in the diagram on the right. All other forces are also shown inthis diagram.FoomaFoonaFcon AThe table below presents the respective horizontal and vertical components of these forcesand the resultant electric force.ForceHorizontal ComponentVertical ComponentFoonA = 0.05625 N–0.05625 NFconA=0.05625 N–0.05625 NFoona = 0.02813 N+0.02813 N (cos 45°) =+0.0199-0.02813 N (sin 45°)=-0.0199 NResultant electric force FEF,=-0.03635 NEF,=-0.07615 NSolving for the magnitude of the resultant electric force,F =V(-0.03635 N)' +(-0.07615 N)²= 0.0844 N.Solving for the direction,|EF,0 = tanEFTA--0.07615 N64,5°= tan|-0.03635 N= 64.5°.Thus, the resultant force is 0.0844 N directed downat 64.5° with the negative x-axis as shown.

Accepted Answer

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3. Suppose that the charges in Sample Problems 1.4 item number 3 are situated as shown. Find the resultant electric force on the charge at A. +q b. 1.00 m A В 1.00 m D b. +9