Current Attempt in Progress Three particles are fixed on an x axis. Particle 1 of charge
is at x = -a and particle 2 of charge
is at x = +a. If their net electrostatic force on particle 3 of charge Q is to be zero, what must be the ratio
when particle 3 is at (a) x = +0.544a and (b) x = +2.07a? (a) Number Units (b) Number Units
The Correct Answer and Explanation is:
We are given three particles along the x-axis:
- Particle 1 has charge q1q_1q1 at position x=−ax = -ax=−a
- Particle 2 has charge q2q_2q2 at position x=+ax = +ax=+a
- Particle 3 has charge QQQ at position x=x3x = x_3x=x3, where x3=+0.544ax_3 = +0.544ax3=+0.544a in part (a), and x3=+2.07ax_3 = +2.07ax3=+2.07a in part (b)
We are asked to find the ratio r=q1q2r = \frac{q_1}{q_2}r=q2q1 such that the net electrostatic force on particle 3 is zero.
Step-by-Step Process
The electrostatic force on a charge QQQ from another charge qqq is given by Coulomb’s Law:F=k∣qQ∣r2F = k \frac{|qQ|}{r^2}F=kr2∣qQ∣
Direction matters: attractive if opposite signs, repulsive if same signs.
Let us assume all charges are positive for simplicity. The direction of each force will then be away from the other charges.
Let the distance from particle 1 to particle 3 be:r1=x3+ar_1 = x_3 + ar1=x3+a
Let the distance from particle 2 to particle 3 be:r2=x3−ar_2 = x_3 – ar2=x3−a
Force from particle 1 on particle 3:F1=kq1Q(x3+a)2F_1 = k \frac{q_1 Q}{(x_3 + a)^2}F1=k(x3+a)2q1Q
Force from particle 2 on particle 3:F2=kq2Q(x3−a)2F_2 = k \frac{q_2 Q}{(x_3 – a)^2}F2=k(x3−a)2q2Q
To make net force zero:F1=F2⇒q1(x3+a)2=q2(x3−a)2⇒q1q2=(x3+ax3−a)2F_1 = F_2 \Rightarrow \frac{q_1}{(x_3 + a)^2} = \frac{q_2}{(x_3 – a)^2} \Rightarrow \frac{q_1}{q_2} = \left( \frac{x_3 + a}{x_3 – a} \right)^2F1=F2⇒(x3+a)2q1=(x3−a)2q2⇒q2q1=(x3−ax3+a)2
Let us now plug in values.
(a) When x3=0.544ax_3 = 0.544ax3=0.544a
q1q2=(0.544a+a0.544a−a)2=(1.544a−0.456a)2=(−3.386)2=11.47\frac{q_1}{q_2} = \left( \frac{0.544a + a}{0.544a – a} \right)^2 = \left( \frac{1.544a}{-0.456a} \right)^2 = \left( -3.386 \right)^2 = 11.47q2q1=(0.544a−a0.544a+a)2=(−0.456a1.544a)2=(−3.386)2=11.47
Answer (a):11.47 (unitless)\boxed{11.47 \ (\text{unitless})}11.47 (unitless)
(b) When x3=2.07ax_3 = 2.07ax3=2.07a
q1q2=(2.07a+a2.07a−a)2=(3.07a1.07a)2=(2.869)2=8.24\frac{q_1}{q_2} = \left( \frac{2.07a + a}{2.07a – a} \right)^2 = \left( \frac{3.07a}{1.07a} \right)^2 = \left( 2.869 \right)^2 = 8.24q2q1=(2.07a−a2.07a+a)2=(1.07a3.07a)2=(2.869)2=8.24
Answer (b):8.24 (unitless)\boxed{8.24 \ (\text{unitless})}8.24 (unitless)
Final Notes:
- These ratios show how the charges must be distributed for particle 3 to experience no net force.
- The closer particle 3 is to one charge, the smaller that charge must be to balance the force from the farther one.
- The ratios are unitless because we’re comparing charge magnitudes.
