Question

In: Physics

A charge Q1 = +4.2 μC is fixed along the y-axis at a distance d =...

A charge Q1 = +4.2 μC is fixed along the y-axis at a distance d = 1.4 m below the origin. A second charge Q2 = +1.6 μC is placed at at a distance d = 1.4 m above the origin as shown.



1)

What is the magnitude of the net force on Q2 due to Q1?

0.0202 N

0.00293 N

0.0077 N

0.0308 N

Your submissions:

  • C

Submitted:

Monday, August 31 at 5:48 PM

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2)

What is the direction of the net force on Q2 due to Q1?

up

down

neither

3)

Now a 3rd charge, Q3 = -4.2 µC (the same magnitude as Q1), is placed a distance 4d = 5.6 m above the origin. How will the magnitude of the net force on Q2 change?

The net force on Q2 will increase.

The net force on Q2 will decrease.

The net force on Q2 will not change.

4)

The 3rd charge can now be changed to any value. In order to make the net force on Q2 equal zero, what must be true about the sign of Q3?

Q3 must be positive.

Q3 must be negative.

No value of Q3 will make the net force on Q2 equal zero.

5)

In order to make the net force on Q2 equal zero, what must be true about the magnitude of Q3?

|Q3|=|Q1||Q3|=|Q1|

|Q3|=32|Q1||Q3|=32|Q1|

|Q3|=94|Q1||Q3|=94|Q1|

|Q3|=4|Q1||Q3|=4|Q1|

|Q3|=16|Q1||Q3|=16|Q1|

No value of Q3 will make the net force on Q2 equal zero.

Solutions

Expert Solution


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