A charge q = 2.4 µC is placed at the point P shown below. What is the force on q (in N)? (Assume the positive direction is to the right. Indicate the direction with the sign of your answer.)

The magnitude of force between two charges q_1\ and\ q_2 separated by a distance r is

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

where k=9×109Nm2/C2

The magnitude of force on charge q=2.7μC=2.7×10-6C by charge 1.5μC=1.5×10-6C which is 2.4+1.2=3.6m away from point P is

F_1=\frac{9\times10^9\times|2.7\times 10^{-6}|\times|1.5\times 10^{-6}|}{3.6^2}

F_1=2.813\times 10^{-3}\ N

The magnitude of force on q by -3.5μC charge which is 1.2m away is

F_2=\ frac {9\times10^9\times|2.7\times 10^{-6}|\times|-3.5\times 10^{-6}|}{1.2^2}

F_2=59.063\times 10^{-3}\ N

We know that like charges repel each other and unlike charges attract each other. q is repelled by 1.5μC and attracted by -3.5μC.

F_1 points to the right it is positive, F_2 points to the left it is negative. The net force on q is

F=+F_1-F_2

F=2.813\times 10^{-3}\ N-59.063\times 10^{-3}\ N

F=-56.25\times10^{-3}\ N

In two significant figures the answer is

F= -0.056  N

Explanation:

explanation:

the masses of the objects and the distance between them

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be