Three charges (q_1 = 6.1 mu C, q_2 = 4 mu C, and q_3 = 1muC) are located at the
ID: 1543312 • Letter: T
Question
Three charges (q_1 = 6.1 mu C, q_2 = 4 mu C, and q_3 = 1muC) are located at the vertices of an equilateral triangle with side d = 6 cm as shown. What is F_3, x, the value of the x-component of the net force on q_3? ____ N You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. What is F_3, y, the value of the y component of the net force on q_3? _____ N You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. A charge q_4 = 1 mu C is now added as shown. What is F_2.x, the x-component of the new net force on q_2? ____ N You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question. What is F_2, y, the y-component of the new net force on q_2?Explanation / Answer
1) F31,x = k q1 q3 cos(60) / d2
F32,x = k q2 q3 cos(60) / d2
q1 , q2 and q3 are magnitude of charges, k is constant 9 x109
F3,x = F31,x + F32,x = k (q1+q2) q3 cos(60) / d2
= 1.26 x 107 N
2) F3,y = F31,y + F32,y
= k (q1-q2) q3 sin(60) / d2 ( F31,y and F21,y are along +ve and -ve direction of y axis, hence (q1-q2) )
= 4.54x106 N
3) F2,x = F21,x + F23,x+ F24,x
= - k ( q3 +q4) q2 cos60 /d2 - k q1 q2 /d2
= - 7.13 x 107 N
4) F2,y = F21,y + F23,y+ F24,y
= 0 + k ( q3 -q4) q2 sin60 /d2 (y component of force of charge 1 is zero)
= 0 ( as q3 = q4)
5) F1,x = F12,x + F13,x+ F14,x
= k q1 q2 /d2 - k ( q3 +q4) q1 cos60 /d2
= 4.5x107 N
6) force on 2 in y direction is zero and in x direction it is along -ve direction. To make it equal to zero sign of 1 should be changed and magnitude should be decreased as , magnitude of force of 3 and 4 on q2 is greater than that of q1
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