A point charge q 1 = -3.5 C is located at the origin of a co-ordinate system. An

Question

A point charge q1 = -3.5 C is located at the origin of a co-ordinate system. Another point charge q2 = 6.6 C is located along the x-axis at a distance x2 = 9.3 cm from q1.

1)

What is F12,x, the value of the x-component of the force that q1 exerts on q2?

2) Charge q2 is now displaced a distance y2 = 2.7 cm in the positive y-direction. What is the new value for the x-component of the force that q1 exerts on q2?

3) A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 7.39 N and points away from q1 and q3. What is the value (sign and magnitude) of the charge q3?

4) How would you change q1 (keeping q2 and q3 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.

5) How would you change q3 (keeping q1 and q2 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 q3 that will result in the fet force on q2 being equal to zero.

Explanation / Answer

Q1.

as q1 and q2 are of opposite sign, force is attractive in nature.

hence force on q2 due to q1 wil be towards q1.

hence the force is directed along -ve x axis.

magnitude =9*10^9*q1*q2/distance^2

=9*10^9*3.5*10^(-6)*6.6*10^(-6)/0.093^2=24.037 N

hence the x component is 24.037 N and along -ve x axis.

Q2.

new coordinate of q2=(9.3,2.7) cm

vector along the force=(0,0)-(9.3,-2.7)=(-9.3,-2.7) cm

distance=sqrt(9.3^2+2.7^2)=9.684 cm

then unit vector along the force=(-9.3,-2.7)/9.684=(-0.96035,-0.27881)

force magnitude=9*10^9*3.5*10^(-6)*6.6*10^(-6)/(9.684*0.01)^2=22.169 N

in component form, total force=22.169*(-0.96035,-0.27881) N

hence x component has a magnitude of 22.169*0.96035=21.29 N and along -ve x axis.

Q3. as net force is away from q1 and q3, charge q3 must be positive so force is repulsive in nature

then q3 is same sign as q2.

net force magnitude=9*10^9*6.6*10^(-6)*((q3/(0.5*9.684*0.01)^2) - (q1/(9.684*0.01)^2))

=594000*(426.53*q3-3.7321*10^(-4))

==>7.39=594000*(426.53*q3-3.7321*10^(-4))

==>426.53*q3=3.8566*10^(-4)

==>q3=9.0417*10^(-7) C

hence q3 is positive and of magnitude 9.0417*10^(-7) C

Q4.

have to increase magnitude of q1.

hence third option is correct.

Q5.have to decrease magnitude of q2.

hence fourth option is correct.

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