Three charges are fixed in the x−y plane as follows: 1.7nC at the origin (0, 0); 2.3nC at (0.60m , 0); –2.0nC at (0, 1.20m ).

Question

Find the force acting on the negative charge.

Express your answer using two significant figures. Enter the x and y components of the force separated by a comma

Answer 1

To find the net force acting on the negative charge, we'll first find the individual forces due to the other two charges and then add them vectorially.

Let's denote the three charges as q1 (1.7 nC), q2 (2.3 nC), and q3 (-2.0 nC), and their positions as r1 (0, 0), r2 (0.60 m, 0), and r3 (0, 1.20 m) respectively.

First, we'll find the force due to q1 on q3. The distance between q1 and q3 is the same as the y-coordinate of q3, which is 1.20 m. So, the magnitude of the force F13 is given by Coulomb's Law:

F13 = k * |q1*q3| / r13^2
where k is Coulomb's constant (8.99 × 10^9 N m^2/C^2), and r13 is the distance between q1 and q3.

F13 = (8.99 × 10^9) * (1.7 × 10^-9 * 2.0 × 10^-9) / (1.20)^2
F13 = 5.12 × 10^-8 N

The direction of this force is along the negative y-axis, as q1 is positively charged and q3 is negatively charged. So, F13 = (0, -5.12 × 10^-8) N.

Next, we'll find the force due to q2 on q3. The distance between q2 and q3 is given by:

r23 = sqrt((0.60)^2 + (1.20)^2) = sqrt(1.80) = 1.34 m

The magnitude of the force F23 is given by Coulomb's Law:

F23 = k * |q2*q3| / r23^2

F23 = (8.99 × 10^9) * (2.3 × 10^-9 * 2.0 × 10^-9) / (1.34)^2
F23 = 1.74 × 10^-8 N

The direction of this force is given by the angle theta between the line connecting q2 and q3 and the negative y-axis:

theta = arctan((0.60)/(1.20)) = arctan(0.5) = 26.57°

The x and y components of F23 are:

F23x = F23 * sin(theta) = 1.74 × 10^-8 * sin(26.57°) = 7.66 × 10^-9 N
F23y = -F23 * cos(theta) = -1.74 × 10^-8 * cos(26.57°) = -1.54 × 10^-8 N
So, F23 = (7.66 × 10^-9, -1.54 × 10^-8) N.

Finally, we'll find the net force acting on q3 by adding the individual forces vectorially:

Fx = F13x + F23x = 0 + 7.66 × 10^-9 = 7.66 × 10^-9 N
Fy = F13y + F23y = -5.12 × 10^-8 + (-1.54 × 10^-8) = -6.66 × 10^-8 N

So, the force acting on the negative charge is approximately (7.66 × 10^-9, -6.66 × 10^-8) N.