Coulomb’s law tells us that the magnitude of the electric force is proportional to the magnitude of each charge, so the magnitude of the electric force quadruples (increases fourfold) when each charge is doubled. It also tells us that the magnitude of the electric force is inversely proportional to the square of the separation distance, so the magnitude of the force reduces one fourth when the separation distance is doubled. The two changes cancel each other and the net change is null; the magnitude of the force is $unchanged$ when each charge $and$ the separation distance are doubled. We can verify this result mathematically by applying these changes, $q_1:;q_1rightarrow q’_1=2q_1$, $q_2:;q_2rightarrow q’_2 = 2q’_2$, $r:;rrightarrow r’ = 2r$, to Coulomb’s law:

$$
F = kfrac{|q_1||q_2|}{r^2}:;Frightarrow F’ = kfrac{|q’_1||q’_2|}{r’^2} = kfrac{|2q_1||2q_2|}{(2r)^2} = F
$$

(the magnitude of the electric force remains the same).

To rank the four systems in order of increasing magnitude of the electric force, we use Coulomb’s law. If we define $F = k|q_1||q_2|/r^2$, then

$$
begin{align*}
text{For system A}:qquad F_{rm A} &= kfrac{|q_1||2q_2|}{r^2} = 2F\
text{For system B}:qquad F_{rm B} &= kfrac{|-2q_1||4q_2|}{(3r)^2} = frac{8}{9}F\
text{For system C}:qquad F_{rm C} &= kfrac{|5q_1||-5q_2|}{(5r)^2} = F\
text{For system D}:qquad F_{rm D} &= kfrac{|16q_1||3q_2|}{(4r)^2} = 3F\
end{align*}
$$

Hence, we rank the systems as $B < C < A < D$.

$$
B < C < A < D
$$

The force exerted on the charge $10q$ has a magnitude that is $equal$ to $F$. Newton’s third law implies that paired action-reaction forces are always equal in magnitude and opposite in direction.

Equal to $F$

(a)

According to Coulomb’s law, the electric force between two charged particles is

$F = k dfrac{|q_1||q_2|}{r^2}$

where $k$ is the proportionality constant and its value is given by

$k = 8.99 times 10^9 :N.m^2/C^2$.

The magnitude of the electric charges is given as

$|q_1| = 0.25:C$ and $|q_2| = 0.11:C$.

The separation between the charges is given as $r =0.88:m$.

So,

$F = 8.99 times 10^9 times dfrac{0.25 times 0.11}{(0.88)^2} = 8.99 times 0.0355 times 10^9 = 3.19 times 10^8:N$

(a) $3.19times10^8;mathrm{N}$
(b) Decreases.

The magnitude of electric force between two point charges is given by Coulomb’s law. Solving for $q$ and substituting the known values of $F$ and $r$, we find

$$
begin{align*}
F &= kfrac{|q_1||q_2|}{r^2} = kfrac{|q||-2q|}{r^2}\
q &= rsqrt{frac{F}{2k}}\
q &= (1.4;mathrm{m})timessqrt{frac{2.2;mathrm{N}}{2(8.99times10^9;mathrm{N}cdotmathrm{m}^2/mathrm{C^2})}}\
&= 15;mumathrm{C}
end{align*}
$$

$15;mu$C