Three charges (q 1 = 5.4 C, q 2 = -5 C, and q 3 = 2.6C) are located at the verti

Question

Three charges (q1 = 5.4 C, q2 = -5 C, and q3 = 2.6C) are located at the vertices of an equilateral triangle with side d = 8.2 cm as shown.

1)

What is F3,x, the value of the x-component of the net force on q3?N

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2)

What is F3,y, the value of the y-component of the net force on q3?N

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3)

A charge q4 = 2.6 C is now added as shown.

What is F2,x, the x-component of the new net force on q2?

N

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4)

What is F2,y, the y-component of the new net force on q2?

N

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5)

What is F1,x, the x-component of the new net force on q1?

N

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6)

How would you change q1 (keeping q2,q3 and q4 fixed) in order to make the net force on q2 equal to zero?

Increase its magnitude and change its sign

Decrease its magnitude and change its sign

Increase its magnitude and keep its sign the same

Decrease its magnitude and keep its sign the same

There is no change you can make to q1 that will result in the fet force on q2 being equal to zero.

Explanation / Answer

1) q1 will repel q3 whether q2 will attract q3.

F3x = [kq2q3 / d^2]cos60 + [kq1q3/d^2]cos60

where k = 9x10^9

q2 = 5 x 10^-6 C , q1 = 5.4 x 10^-6 C , q3 = 2.6 x 10^-6 C

d = 8.2 cm = 0.082 m

F3x = 18.10 N

2) F3y = - [kq2q3 / d^2]sin60 + [kq1q3/d^2]sin60

      = 1.21 N

3) F2x = - [kq1q2/ d^2 ] - [kq2q3/d^2]cos60 - [kq2q4/d^2]cos60

     = - 47.56 N

4) q3 and q4 have same charges. so

F2y = [kq2q3/d^2]sin60 - [kq2q4/d^2]sin60 = 0

5) F1x = [kq1q2/ d^2 ] - [kq1q3/d^2]cos60 - [kq1q4/d^2]cos60

     = 15.41 N

6)F2x = - [kq1q2/ d^2 ] - [kq2q3/d^2]cos60 - [kq2q4/d^2]cos60 = 0

first we have to change the sign of q1.

now F2x = [kq1q2/ d^2 ] - [kq2q3/d^2]cos60 - [kq2q4/d^2]cos60 = 0

q1 = q3cos60 + q4 cos60 = 2(2.6cos60) = 2.6 uC

but q1 is 5.4uC .

so decrease the magnitude and change the sign.

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