Question

A point charge q1 = 3.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 10 m.

Answer 1

Answer:

a) 1.6 mN  b) -1.6 mN  c) -1.6 mN  d) 1.6 mN

Explanation:

The electrostatic force between 2 point charges, obeys the Coulomb's Law, that can be expressed as follows:

F₁₂ = k*q₁*q₂/(r₁₂)² (in magnitude)

The direction of the force, is along the  line that joins the  charges (along the x axis) and as q₁ and q₂ are of the same sign, aims away from both charges.

a) So, for the force on q₂, we have:

F₁₂ = 9*18*10⁻⁵ N = 1.6 mN (positive as it is aiming in the positive x direction)

b) The force on q1, according to Newton's 3rd Law, is just equal and opposite to the one on q2:

F₂₁ = (-9*18*10⁻⁵) N = -1.6 mN (towards the negative x direction, away from q1)

c) If q₂ were -6.0 μC, the force will be the same in magnitude, but as now both charges have different signs, they wil attract each other, so the direction of the forces will be exactly the opposite to the first case:

F₁₂ = -1.6 mN (going towards the origin, where q₁ is located)

F₂₁ =  1.6 mN (going in the positive x direction, towards q₂)

Answer 2

Answer:

a. $F_{21}=(1.62*10^{-3}N)i$

b.  $F_{12}=-(1.62*10^{-3}N)i$

c. $F_{21}=-(1.62*10^{-3}N)i$,  $F_{12}=(1.62*10^{-3}N)i$

Explanation:

Recall that like charged particles repel while unlike charges attract. hence we can conclude that both particles will repel with a force of magnitude

$F=\frac{kq_{1}q_{2}}{r^{2}}$

Where $k=9*10^{9}Nm^{2}C^{-2}\\ r=10m\\q_{2}=6.0*10^{-6}C\\q_{1}=3.0*10^{-6}C$.

Substituting values into the above equation we have

$F=\frac{9*10^{9}*3.0*10^{-6}*6.0*10^{-6}}{10^{2}} \\F=\frac{162*10^{(9-6-6)}}{100}\\F=\frac{162*10^{(-3)}}{100}\\\\F=\frac{0.162}{100}\\\\F=1.62*10^{-3}N$

Hence the electric force on charge q2 is of magnitude $F=1.62*10^{-3}N$  and the direction is away from q2 in the positive x direction  which can be represented as a vector form as

$F=(1.62*10^{-3}N)i$.

bTo determine the electric force on charge q1, the magnitude remain the same

$F=\frac{9*10^{9}*3.0*10^{-6}*6.0*10^{-6}}{10^{2}} \\F=\frac{162*10^{(9-6-6)}}{100}\\F=\frac{162*10^{(-3)}}{100}\\\\F=\frac{0.162}{100}\\\\F=1.62*10^{-3}N$.. but in the opposite direction which is the negative x-axis

$F=-(1.62*10^{-3}N)i$

c.if the q2 is -6.0uc, the magnitude of the force will be the same but the direction will be different since both charges are unlike, hence the electric force on q 2 as in (a) above will be towards q1  which can be represented as a vector form as

$F=-(1.62*10^{-3}N)i$ and for the second case scenario the forc will be directed towards the second charge which can be represented as  $F=(1.62*10^{-3}N)i$