Question

This is one I don’t understand ? How do you calculate formula ? Can you show all step

Answer 1

Answer:

Net force =  13.36 N

Explanation:

Recall the formula for the force between two charges q1 ans q2 separated a distance "d" based on Coulomb's Law:

$F_C=k\,\frac{q_1\,*\,q_2}{d^2}$

so, we need to calculate the force exerted on the charge 10 micro Coulombs by the 20 micro-Coulombs charge as they are separated 0.8 meters (haf the distance of the original two charges.

Notice that charges q1 and q3 are both positive charges so the force we obtain will be pointing in the direction towards charge q2. The magnitude of this is:

$F_{13}=k\,\frac{20\,\,10^{-6}\,*\,10\,\,10^{-6}}{0.8^2}=9\,\,10^{9}\,\frac{200\,\,10^{-12}}{0.8^2}= 2812.5 \,\,10^{-3} \,N = 2.8125 \,\,N$

We can calculate now the force exerted by charge q2 on charge q3 using the same reasoning, and noticing that since charge q2 is negative, the force on q3 will be of attraction, then also pointing in the direction towards q2:

$F_{23}=k\,\frac{75\,\,10^{-6}\,*\,10\,\,10^{-6}}{0.8^2}=9\,\,10^{9}\,\frac{750\,\,10^{-12}}{0.8^2}= 10546.875 \,\,10^{-3} \,N = 10.547 \,\,N$

Now, we need to add both forces to get the final force (recall they are both in the same direction):

Net force = 10.547 N + 2.813 N = 13.36 N