(13%) Problem 4: In the figure, the point charges are located at the corners of

Question

(13%) Problem 4: In the figure, the point charges are located at the corners of an equilateral triangle 29 cm on a side ga qc ©theexpertta.com 25% Part (a) Find the magnitude of the electric field in N/C at the location of qa, given that q,-70.8 .C and q.--5.9 pC Correct Answer lEa1002000 Final Grade 100% Student Answer lEal = 1.0 * 106 El 1000000 Grade Detail Correct Student Feedback Final Answer Credit 100% Submission History Answer Hints Feedback Totals 0% E I 10106 lEal = 1000000 0% 0% 0% Totals 0% 0% 0% 0% 25% Part (b) Find the direction of the electric field at qa in degrees above the negative x-axis with origin at qa 25% Part (c) What is the magnitude of the force in N on qa, given that q,-12 nC? 25% Part (d) What is the direction of the force on qa in degrees above the negative x-axis with origin at

Explanation / Answer

Given

charges are Qa = 1.2nC, Qb = 10.8 nC, and Qc = 5.9 nC

(a) Find the electric field at the location of Qa, due to Qb,Qc

we knwo that the electric field due a point charge q at a distance r is

E = kq/r^2

so here the electric field at Qa due to Qb is , first qb = 10.8*10^-9 C and the distance r = 0.29 m

Eb = 9*10^9*(10.8*10^-9)/(0.29)^2 N/C = 1155.766944 N/C

the electric field at the location of Qa, due to Qc.

Qc = -5.9 nC , r = 29 cm = 0.29 m

Ec = 9*10^9*(-5.9*10^-9)/(0.29)^2 N/C = -631.39120 N/C

the net electric field at the location of Qa due to Qb,Qc

writing the components of the elelctric fields of Eb,Ec

Eb = Ebx i + Eby j

Ec = Ecx i + Ecy j

Eb = Eb cos (120) i + Eb sin (120) j = 1155.766944 cos 135 i +1155.766944 sin 135 j

Eb = -817.250644 N/C i + 817.250644 N/C j

Ec = Ec cos 240 i + Ec sin 240 j = 631.39120 cos 240 N/C i + -631.39120 sin30 N/C

Ec = -315.6956 N/C i - 546.80082 N/C j

the net field is  

Ea = (-817.250644 -315.6956 ) i +(817.250644 - 546.80082 ) j

Ea = -1132946.244 i + 270449.824 j

the magnitude is

E3 = sqrt((-1132.946244)^2+(270.449824 )^2) N/C = 1164.7791 N/C

pART (b)

the direction is theta arc tan ((270.449824)/(-1132.946244))

theta = -13.43 degrees = 360-13.43 = 346.57 degrees with +x axis  

the angle above the -ve x axis is with origin at Qa is 346.57-180 = 166.57 degrees

Part (C)

Find the electric force at Qa, due to Qb,Qc

we knwo that the electric force from coulomb's law

F = kq1*q2/r^2

so here the electric force at Qa due to Qb is , first qb = 10.8*10^-9 C and the distance r = 0.29 m

Fb = 9*10^9*(10.8*10^-9)(1.2*10^-9)/(0.29)^2 N/C = 1.38692*10^-6 N

and Fc = 9*10^9*(-5.9*10^-9)(1.2*10^-9)/(0.29)^2 N/C = (-7.5766944)*10^-7 N

the net electric force at Qa due to Qb,Qc

writing the components of the elelctric fields of Eb,Ec

Fb = Fbx i + Fby j

Fc = Fcx i + Fcy j

Fb = Fb cos (120) i + Fb sin (120) j = 1.38692*10^-6 cos 135 i +1.38692*10^-6 sin 135 j

Fb = (-9.8070)*10^-7 )N i + 9.8070*10^-7 N j

Fc = Fc cos 240 i + Fc sin 240 j = (-7.5766944)*10^-7 cos 240 N i + (-7.5766944)*10^-7 sin30 N

Fc = 3.7883472*10^-7 N i -3.7883472*10^-7 N j

the net field is  

Fa = (-9.8070*10^-7 +3.7883472*10^-7 ) i +(9.8070*10^-7 -3.7883472*10^-7 ) j

Fa = (-6.0186528)*10^-7 i +6.0186528*10^-7 j

the magnitude is

E3 = sqrt((-6.0186528)*10^-7)^2+(6.0186528*10^-7 )^2) N = 8.5116604169748*10^-7 N

Part (d)

the direction is theta arc tan ((6.0186528*10^-7)/(-6.0186528)10^-7))

theta = -45 degrees = 315-180 = 135 degrees

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