Question

Two small plastic spheres are given positive electrical charges. When they are 15.0 cm apart, the repulsive force between them has magnitude 0.220 N. What is the charge on each sphere?
(a) if the two charges are equal?
(b) if one sphere has four times the charge of the other?

Answer 1

Answer:

a) q_1=q_2= 7.42*10^-7 C

b) q_2= 3.7102*10^-7 C , q_1 = 14.8*10^-7 C

Explanation:

Given:

F_e = 0.220 N

separation between spheres r = 0.15 m

Electrostatic constant k = 8.99*10^9

Find: charge on each sphere

part a

q_1 = q_2

Using coulomb's law:

F_e = k*q_1*q_2 / r^2

q_1^2 = F_e*r^2/k

q_1=q_2= sqrt (F_e*r^2/k)

Plug in the values and evaluate:

q_1=q_2= sqrt (0.22*0.15^2/8.99*10^9)

q_1=q_2= 7.42*10^-7 C

part b

q_1 = 4*q_2

Using coulomb's law:

F_e = k*q_1*q_2 / r^2

q_2^2 = F_e*r^2/4*k

q_2= sqrt (F_e*r^2/4*k)

Plug in the values and evaluate:

q_2= sqrt (0.22*0.15^2/4*8.99*10^9)

q_2= 3.7102*10^-7 C

q_1 = 14.8*10^-7 C

Answer 2

Answer:

a) 0.74 μC b) q1 = 0.37 μC q₂ = 1.48 μC

Explanation:

Assuming that the spheres are small enough so both can be treated as point charges, the repulsive force between them must obey Coulomb's Law, as follows:

$F=\frac{k*q1*q2}{d^{2}}$

a) If the two charges are equal , q₁ = q₂ = q, then:

$F=\frac{k*q^{2} }{d^{2}}$

where k= 9*10⁹ N*m²/C², F= 0.220 N and d =0.15 m

Replacing in the above equation and solving for q, we have:

 $F=\frac{k*q^{2} }{d^{2}} = F=\frac{9*10(9)(N*m2/C2)*q^{2}}{(0.15m)^{2}} =0.22 N$

⇒  $q =\sqrt{\frac{(0.22N*(0.15m)^{2} x}{9*10(9)N*m2/C2}} =0.74 uC$

 ⇒ q₁ = q₂ = q = 0.74 μC

b) All the same considerations apply, the only difference is that for this case, q₁ = q and q₂ = 4*q.

The expression for the electrostatic force is now:

 $F=\frac{k*q^{2} }{d^{2}} = F=\frac{9*10(9)(N*m2/C2)*4q^{2}}{(0.15m)^{2}} =0.22 N$

Solving for q;

 $q =\sqrt{\frac{(0.22N*(0.15m)^{2} x}{4*9*10(9)N*m2/C2}} =0.37 uC$

so, q₁= 0.37 μC ⇒ q₂ = 4*q₁ = 0.37 μC * 4 = 1.48 μC