Steps for Force problems 1.) draw picture - label all charges (q1,q2,q3) label a

Question

Steps for Force problems

1.) draw picture - label all charges (q1,q2,q3)

label all r ( draw a straight line between each pair of charges, label with appropriate subscript ( r12 : from q1 to q2)

2. draw force arrows between the charge in question and each of the other charges, arrows should be attacting ( pointing towards each other) if the charges are opposite, and point away if repelling )

label according to Fxy the force by (x) on (y). F32 is the force by q3 on q2)

3. isolate the chrage in question - draw it on its own x-y axis and transer only the force arrows on that charge. now is a good time to use geometry to find any angles if you havent

4. Find the magnitude of all your force arrows

Explanation / Answer

rp = - 3 i - 3 j

r1 = - 2 i + 3 j

r2 = 3 i + 2 j

r3 = 2 i - 2 j

(rp - r1) = ( - 3 i - 3 j ) - (- 2 i + 3 j) = - i - 6 j

|(rp - r1)| = sqrt((- 1)2 + (- 6)2) = 6.1

(rp - r2) = ( - 3 i - 3 j ) - (3i + 2j) = - 6 i - 5 j

|(rp - r2)| = sqrt((- 6)2 + (- 5)2) = 7.8

(rp - r3) = ( - 3 i - 3 j ) - (2i - 2j) = - 5 i - j

|(rp - r3)| = sqrt((- 5)2 + (- 1)2) = 5.1

E1 = electric field by q1 at P = k q1 (rp - r1)/(|(rp - r1)|)3

E1 = (9 x 109) (- 5 x 10-9) ( - i - 6 j)/(6.1)3 = - (0.198) ( - i - 6 j) = 0.198 i + 1.188 j

E2 = electric field by q2 at P = k q2 (rp - r2)/(|(rp - r2)|)3

E2 = (9 x 109) (8 x 10-9) (- 6 i - 5 j)/(7.8)3 = (0.152) ( - 6 i - 5 j) = 0.912 i - 0.76 j

E3 = electric field by q3 at P = k q3 (rp - r3)/(|(rp - r3)|)3

E3 = - (9 x 109) (2 x 10-9) ( - 5 i - j)/(5.1)3 = - (0.136) ( - 5 i - j) = 0.68 i + 0.136 j

Total electric field at P is given as

E = E1 + E2 + E3

E = 0.198 i + 1.188 j + 0.912 i - 0.76 j + 0.68 i + 0.136 j

E = 1.79 i + 0.564 j

X-component = 1.79 N/C

Y-component = 0.564 N/C

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