Two point charges (q 1 = -4uC and q 2 = 7.9 uC) are fixed along the x-axis, sepa

Question

Two point charges (q1 = -4uC and q2 = 7.9 uC) are fixed along the x-axis, separated by a distance d = 9.8 cm.  Point P is located at (x,y) = (d,d).         

1)

2)

3)

A third point charge q3 =  2.5 uC is now positioned along the y-axis at a distance d = 9.8 cm from q1 as shown.  What is Ex(P), the x-component of the field produced by all 3 charges at point P?

4)

5)

Two point charges (q1 = -4uC and q2 = 7.9 uC) are fixed along the x-axis, separated by a distance d = 9.8 cm. Point P is located at (x,y) = (d,d). What is Ex(P), the value of the x-component of the electric field produced by q1 and q2 at point P? What is Ey(P), the value of the y-component of the electric field produced by q1 and q2 at point P? A third point charge q3 = 2.5 uC is now positioned along the y-axis at a distance d = 9.8 cm from q1 as shown. What is Ex(P), the x-component of the field produced by all 3 charges at point P? Suppose all charges are now doubled (i.e., q1 = -8 ?C, q2 = 15.8 uC, q3 = 5 uC), how will the electric field at P change? Its magnitude will increase by less than a factor of two and its direction will remain the same Its magnitude will increase by less than a factor of two and its direction will change Its magnitude will double and its direction will remain the same Its magnitude will double and its direction will change How would you change q1 (keeping q2 and q3 fixed) in order to make the electric field at point P 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 electric field at point P being equal to zero.

Explanation / Answer

1) only 1makes a field in the x

Ex = - E1 cos(45) =-9E9*4.0E-6/(2*9.8E-2^2)*cos(45 degrees)=-1.325E6

2) Ey = -E1 sin(45) + E2 = -9E9*4.0E-6/(2*9.8E-2^2)sin(45 degrees)+9E9*7.9E-6/(9.8E-2^2)=6.078E6

3) on q3 and q1 make field in x direction

Ex = - E1 cos(45) + E3 = -9E9*4.0E-6/(2*9.8E-2^2)*cos(45 degrees)+9E9*2.5E-6/(9.8E-2^2)=1.018E6

4) it will double and direction will be the same so C)

5) E)

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