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.

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

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

2)

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)

F1 , x = 9 X 109 X (q1q2 / d2) - [(q3 + q4 ) X q1) / d2 X cos]

but here = 600

F1 , x = 9 X 109 X (q1q2 / d2) - [(q3 + q4 ) X q1) / d2 X cos60]

= 9 X 109 X (q1q2 / d2) - [(q3 + q4 ) X q1) / d2 X 1/2]
= 9 X 109 X ( q1/d2) X [ (q2-(q3 + q4) / 2 ]
= 9 X 109 X ( 5.4 X 10-6 / (0.082)2) X [ (5 X 10-6 )-( 2.6 X 10-6+ 2.6 X 10-6) / 2 ]
= 9 X 103 X ( 5.4 / 6.724 X 10-3 ) X [ (5.0) - (5.2) / 2 ]

= 9 X 103 X ( 5.4 / 6.724 X 10-3 ) X [ (0.2 / 2 ]

= 9 X 103 X ( 5.4 / 6.724 X 10-3 ) X 0.1

= 9 X (0.803) X 106 X 0.1

= 0.7227 X 106 N

2)

Decrease its magnitude and change its sign

and

q2 and q3 are decrease because force in the negative direction then decreases the charge q1

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