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In Example 21.4, suppose the point charge on the y-axis at y=-0.30m has negative charge -2μC, and the other charges remain the same. Find the magnitude and direction of the net force on Q. How does your answer differ from that in Example 21.4? Explain the differences.

Short Answer

Expert verified

The net force is 0.35 N and its direction is in the negative y-direction.

Step by step solution

01

Coulomb’s law

Coulomb's law states that the magnitude F of the force exerted by two-point chargesq1 andq2 separated by r is directly proportional to the product of the charges(q1×q2) and inversely proportional to the square of the distance between them.

F=kq1q2r2

Here; F is the force on each point charge exerting on each other, k is the proportionality constant, r is the distance between the chargesq1 and q2.

02

Magnitude and direction of the net force on Q

Consider the given data as below.

The charge,q2=-2.00μC=-2.00×10-6C

The distance between two point charges, r = 0..5 m

The Coulomb’s constant, localid="1665130019862" k=9×109Nm2/C2

In example 21.4, the net force on Q is a repulsive force, but now Q has both repulsive and attractive forces. Therefore, the charge,

Q=4.00μC=4.00×10-6C

Because q1exerted repulsive forces in the negative direction of the y-axis with angle αwith the x-axis, the negative charge q2exerted an attractive force towards itself, so the y-axis direction will be;

Because the two forces acting on Q have the same magnitude, the net force in the new case will be;

Fnet=F1Q+F2Q

Here, localid="1665130545351" F1Q=F2Q. Therefore,

localid="1665130595651" Fnet=2F2Q=2kq2Qr2sinα

Substitute known values in the above expression, and you have

Fnet=29×109N·m2/C2×-3.00×10-6C4.00×10-6C0.5m20.3m0.5m=345.6×10-3N=0.35N

Hence, the net force is 0.35 N and its direction is in the negative y-direction.

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