Question

The attractive electric force between point charges q and -2q has a magnitude of 2.2 N when the separation
between the charges is 1.4 meters. What is the magnitude of charge q?

Answer 1

Answer:

The Magnitude of charge = q = 15.47 micro coulomb  

Step-by-step explanation:

The attractive force between two points q and - 2 q = F = 2.2 Newton

The separation between two charges = 1.4 meters

Now from Coulomb's law = F = $(k \frac{q1 q2}{r^{2}})$  where k= $9 \times 10^{9}$ $\frac{N m^{2}}{c^{2}}$

So , 2.2 = $9 \times 10^{9}$ ( $(\frac{(q) ( - 2q)}{1.4^{2}})$

So , q ² = ($\frac{0.479 \times 10^{-9}}{-2}$)

      For charge magnitude

   I.e q ² = $2.395 \times 10^{-10}$

 so, q =  $1.547 \times 10^{-5}$

      q = 15.47 \times 10^{-6} c

Or,  q = 15.47 micro coulomb

Hence  the Magnitude of charge = q = 15.47 micro coulomb   Answer

Answer 2

By using the electromagnetic force equation, we will see that the magnitude of the charge is q = 1.55*10^5 Coulombs.

How to get the magnitude of the charge?

The force between two charges q₁ and q₂ is:

$|F| = k*\frac{q_1*q_2}{r^2}$

Where r is the distance between the charges, r = 1.4 meters, and k is a constant.

k = 8.99*10^9 kg*m^3/s^2*C^2

In this case, we know that:

q₁ = q

q₂ = -2q

|F| = 2.2 N

Replacing all that we get:

$2.2N = (8.99*10^9 kg*m^3/s^2*C^2)\frac{|q*-2q|}{(1.4m)^2} \\\\2.2N*\frac{(1.4m)^2}{ (8.99*10^9 kg*m^3/s^2*C^2)} = 2q^2\\\\\sqrt{2.2N*\frac{(1.4m)^2}{ 2*(8.99*10^9 kg*m^3/s^2*C^2)}} = q = 1.55*10^-5 C$

So we have that the magnitude of the charge q is 1.55*10^5 Coulombs.

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