Question

Two particles, one with charge -2.93 ?C and one with charge 1.87 ?C, are 3.59 centimeters apart. What is the magnitude of the force that one particle exerts on the other? B. Two new particles, which have identical positive charge q3, are placed the same 3.59 centimeters apart, and the force between them is measured to be the same as that between the original particles. What is q3?

Answer 1

Answer:

(A) F = $3.826\times 10^{13}\ N$

(B) $q_3 = 2.34\ C$

Explanation:

Given:

(A)

Charge on first particle (q₁) = -2.93 C

Charge on second particle (q₂) = 1.87 C

Separation (d) = 3.59 cm = 0.0359 m [∵ 1 cm = 0.01 m]

The magnitude of force is given by Coulomb's law which states that, the magnitude of force acting between two charged particles separated by a distance is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

Therefore, the magnitude of force is given as:

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

Where, $k=9\times 10^9\ N\cdot m^2/ C^2$ is the coulomb's constant.

Plug in the given values and solve for 'F'. This gives,

$F=\frac{9\times 10^9\times 2.93\times 1.87}{(0.0359)^2}\\\\F=3.826\times 10^{13}\ N$

Therefore, the magnitude of the force that one particle exerts on the other is $3.826\times 10^{13}\ N$

(B)

Given:

Magnitude of each charge is $q_3$.

Separation (d) = 3.59 cm = 0.0359 m

Force is same as (A). So, $F=3.826\times 10^{13}\ N$

Now, using Coulomb's law, we have:

$F=\dfrac{k|q_3||q_3|}{d^2}$

Plug in the given values and solve for $q_3$. This gives,

$3.826\times 10^{13}=\frac{9\times 10^9\times q_3^{2}}{(0.0359)^2}\\\\3.826\times 10^{13}=6.983\times 10^{12}\times q_3^{2}\\\\q_3^{2}=\dfrac{3.826\times 10^{13}}{6.983\times 10^{12}}\\\\q_3^{2}=5.479\\\\q_3=\sqrt{5.479}\\\\q_3=2.34\ C$

Therefore, the magnitude of each of the two new charges is 2.34 C.