Three charges are placed at the vertices of an equilateral triangle. The value o
ID: 1586267 • Letter: T
Question
Three charges are placed at the vertices of an equilateral triangle. The value of the charge Q1 is given below. Using the value shown in the picture compute in the given coordinate system 1. The value of the force exerted by Q2 and Q3 on Q1 along the x direction 2. The value of the force exerted by Q2 1.20 m and Q3 on Q1 along the y direction Q2 8.0 C Q3 6.0 AuC 3. T he value of the force exerted by Q and Q3 on Q2 along the x direction 4. The value of the force exerted by Q1 and Q3 on Q2 along the y direction 5. The value of the force exerted by Q1 and Q2 on Q3 along the x direction 6. The value of the force exerted by Q1 and Q2 on Q3 along the y direction The value of the charge e is 7.3291 uC Question #1 Trials l 5 Check 5 Question #2 Trials l Question #3 Trials l 5 Check Question #4 Trials l 5 Check Question #5 Trials l 5 Question #6 Trials l 5Explanation / Answer
1)
F1x = F21x + F31x = -k*Q2*Q1*cos60/r^2 + k*Q3*Q1*cos60/r^2
F1x = -(9*10^9*8*10^-6*7.3291*10^-6*cos60)/1.2^2 + (9*10^9*6*10^-6*7.3291*10^-6*cos60)/1.2^2
F1x = -0.045 N
2)
F1y = F21y + F31y = -k*Q2*Q1*sin60/r^2 - k*Q3*Q1*sin60/r^2
F1y = -(9*10^9*8*10^-6*7.3291*10^-6*sin60)/1.2^2 - (9*10^9*6*10^-6*7.3291*10^-6*sin60)/1.2^2
F1y = -0.56 N
3)
F2x = F12x + F32x = k*Q1*Q2*cos60/r^2 + k*Q3*Q2*cos60/r^2
F2x = +(9*10^9*7.3291*10^-6*8*10^-6*cos60)/1.2^2 - (9*10^9*6*10^-6*8*10^-6*cos60)/1.2^2
F2x = 0.033 N
4)
F2y = F12y + F32y = k*Q1*Q2*sin60/r^2 +0
F2y = +(9*10^9*7.3291*10^-6*8*10^-6*sin60)/1.2^2
F2y = 0.3173 N
5)
F3x = F13x + F23x = -k*Q1*Q3*cos60/r^2 + k*Q2*Q3*cos60/r^2
F3x = -(9*10^9*7.3291*10^-6*6*10^-6*cos60)/1.2^2 + (9*10^9*8*10^-6*6*10^-6*cos60)/1.2^2
F3x = 0.0126 N
6)
F3y = F13y + F23y = k*Q1*Q3*sin60/r^2 +0
F3y = +(9*10^9*7.3291*10^-6*6*10^-6*sin60)/1.2^2
F3y = 0.238 N
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