Object A has a positive charge of +2Q Coulombs. Object B has a positive charge o

Question

Object A has a positive charge of +2Q Coulombs. Object B has a positive charge of +Q coulombs. At which of the points labeled 1, 2, 3, orf 4 could a test charge remain at rest? 1 2 3 4 The answer depends on the test charge, which is not given. A negative 0.3-mu C charge at the origin (lower left in the sketch). An identical charge is placed a distance a = 0.03 m from the origin on the x axis. A third 0.3 mu C charge that is positive is placed 0.03 m from the origin on the y axis. The magnitude and direction (relative to the positive x-axis) of the net force on the charge at the origin are: 20 N, 45 degree 1.3 N, -45 degree 3.9 N, 225 degree 3.9 N, 90 degree 1.3 N, 135 degree A small conducting ball has a mass of 5.0 times 10^-2 kg and a negative charge of -300 mu C. What electric field and direction are needed to exactly balance that weight of the ball? (g = 9.8 m/s^2)?

Explanation / Answer

Q4. let distance from A be d1 and distance from B be d2.

force to be equal and cancelling, the test charge should be placed in between A and B.

equating the forces:

k*2*Q*q/d1^2=k*Q*q/d2^2

==>(d1/d2)^2=2

==>d1/d2=1.414

so the test charge should be closer to B .

hence point 3 is correct answer.

Q5. magnitude of forces due to both the charges are equal.

magnitude of each force=9*10^9*0.3*10^(-6)*0.3*10^(-6)/0.03^2=0.9 N

force due to +0.3 uC is attractive and is along +ve y axis.

force due to -0.3 uC is repulsive and is along -ve x axis

so net force magnitude=sqrt(force along x axis^2+force along y axis^2)

=sqrt(0.9^2+0.9^2)=1.2728 N

direction is 135 degree with +ve x axis.

so option E is correct.

Q6.as charge is negative, force on the charge is in opposite direction to that of electric field

as required electric force is in upward direction, electric field should be in downward direction

equating force magnitudes,

q*E=m*g

==>E=m*g/q=5*10^(-2)*9.8/(300*10^(-6))=1633.33 N/C

so field is 1633.33 N/C, upward direction

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