Question

2. Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m. The charges are +7.0μC, -8.0 μC and -6.0 μC. Calculate the net force on charge 1 due to the other two charges in unit vector notation. Give values for the magnitude and direction of the force, too.

Answer 1

To calculate the net force on charge 1 due to the other two charges, we need to find the individual forces between charge 1 and the other charges and then combine them vectorially using Coulomb's law.

Step-by-step explanation:

To calculate the net force on charge 1 due to the other two charges, we need to find the individual forces between charge 1 and the other charges and then combine them vectorially. The magnitude of the force between two charges can be calculated using Coulomb's law:

F = k * (|q1| * |q2|) / (r^2)

Where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.

Let's calculate the individual forces:

1. Force between charge 1 (+7.0μC) and charge 2 (-8.0 μC) can be calculated using the formula:
   F12 = k * (|q1| * |q2|) / (r^2) = k * (|7.0| * |-8.0|) / (1.20^2)
2. Force between charge 1 (+7.0μC) and charge 3 (-6.0 μC) can be calculated using the same formula:
   F13 = k * (|q1| * |q3|) / (r^2) = k * (|7.0| * |-6.0|) / (1.20^2)

Next, we can calculate the net force on charge 1 by adding the forces vectorially:

Net Force on charge 1 = F12 + F13

Answer 2

0.53 N, 25.6°

Step-by-step explanation:

side of triangle, a = 1.2 m

q = 7 μC

q1 = - 8 μC

q2 = - 6 μC

Let F1 be the force between q and q1

By using the coulomb's law

`F_(1)=(Kq_(1)q)/(a^(2))`

`F_(1)=(9* 10^(9)* 7* 10^(-6)* 8* 10^(-6))/(1.2^(2))`

F1 = 0.35 N

Let F2 be the force between q and q2

By using the coulomb's law

`F_(2)=(Kq_(2)q)/(a^(2))`

`F_(2)=(9* 10^(9)* 7* 10^(-6)* 6* 10^(-6))/(1.2^(2))`

F2 = 0.26 N

Write the forces in the vector form

`\overrightarrow{F_(1)}=0.35\widehat{i}`

`\overrightarrow{F_(2)}=0.26\left ( Cos60 \widehat{i}+Sin60\widehat{j}\right )`

`\overrightarrow{F_(2)}=0.13 \widehat{i}+0.23\widehat{j}`

Net force

`\overrightarrow{F}=\overrightarrow{F_(1)}+\overrightarrow{F_(2)}`

`\overrightarrow{F}=0.48 \widehat{i}+0.23\widehat{j}`

Magnitude of the force

`F=\sqrt{0.48^(2)+0.23^(2)}`

F = 0.53 N

Direction of force with x axis

`tan\theta =(0.23)/(0.48)`

θ = 25.6°