Question

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.3 cm . Two of the particles have a negative charge: q1 = -8.2 nC and q2 = -16.4 nC . The remaining particle has a positive charge, q3 = 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

Answer 1

The net electric force acting on particle 3 due to particle 1 and particle 2 is 4.55 N.

Step-by-step explanation:

To find the net electric force acting on particle 3 due to particle 1 and particle 2, we can use Coulomb's Law, which states that the electric force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is: F = k × (|q1 × q3| / r^2)

Where:

• F is the net electric force
• k is the electrostatic constant
• r is the distance between the charges
• q1 and q3 are the charges of particle 1 and particle 3 respectively

Substituting the given values into the formula:

F = (9 × 10^9 Nm^2/C^2) × (|(-8.2 × 10^-9 C) × (8.0 × 10^-9 C)| / (3.3 × 10^-2 m)^2)

Simplifying the equation, we get:

F = 4.55 N

Therefore, the net electric force acting on particle 3 is 4.55 N.

Answer 2

`1.44* 10^(-3)` N

Step-by-step explanation:

We are given that three charged particle are placed at each corner of equilateral triangle.

`q_1=-8.2 nC,q_2=-16.4 nC,q_3=8.0nC`

`q_1=-8.2* 10^(-9) C`

`q_2=-16.4* 10^(-9) C`

`q_3=8.0* 10^(-9) C`

Side of equilateral triangle =3.3 cm=
`(3.3)/(100)=0.033m`

We know that each angle of equilateral angle=
`60^(\circ)`

Net force=F =
`\sum(kQq )/(d^2)`

Where k=
`9* 10^9 Nm^2/C^2`

If we bisect the angle at
`q_3` then we have 30 degrees from there to either charge.

Direction of vertical force due to charge
`q_1` and
`q_2`

Therefore, force will be added

Vertical force=
`9* 10^9* q_3(q_1+q_2)(cos30)/((0.033)^2))`

Vertical net force=
`9* 10^9* 8* 10^(-9)(-8.2-16.4)* 10^(-9)* 10^6*(\sqrt3)/(2* 1089)`

Vertical force =
`9* 8(-24.6)* 10^(-9)* 10^6* 1.732* (1)/(2178)`

Vertical force=
`-1.41* 10^(-3)N` (towards
`q_1}`

Horizontal component are opposite in direction then will b subtracted

Horizontal force=
`9* 10^9* 8* 10^(-9)(-8.2+16.4) * 10^(-9)* (sin30)/((0.033)^2)`

Horizontal force=
`0.27* 10^-3}` N(towards
`q_2`

Net electric force acting on particle 3 due to particle =
`√(F^2_x+F^2_y)`

Net force=
`\sqrt{(-1.41* 10^(-3))^2+(0.27* 10^(-3))^2}`

Net force=
`1.44* 10^(-3)` N