Question

Find the net force on q3. Include the direction ( +or-).

Answer 1

Answer:

72.16 N

Explanation:

Given:

q₁ = -53.0 μC

q₂ = 105 μC

q₃ = -88.0 μC

q₁ to q₂ = 0.50 m

q₂ to q₃ = 0.95 m

To find:

Net force on q₃

Solution:

First compute net electric field on q₃

E = F/q = k.Q/d²

The formula of electric field at q₃:

E = k.Q / r²

Where    

r is distance

Q is magnitude of charge

k is a constant with a value of 8.99 x 10⁹ N m²/C²

When

q₂ to q₃ = 0.95m and

q₂ = 105 μC then

Find electric field due to charge q₂

E = ( (8.99 x 10⁹)x( 105 x 10⁻⁶) ) / 0.95²

  =  (8990000000  x 0.000105) / 0.9025

  = 943950  / 0.9025

  = 1045927.977839

  = 1.046 x 10⁶ N/C

This interprets that it will act or point away from q₂

q₁ to q₂= 0.50m

q₂ to q₃ = 0.95m and

q₁ = -53 μC then

Find electric field due to charge q₁

E = (8.99 x 10⁹) x (53 x 10⁻⁶) / (0 .50 + 0.95)²

  =  (8990000000  x  0.000053) / (1.45)²

  = 476470 /2.1025

  = 226620.689655

  = 0.227  x 10⁶ N/C

This interprets that it will act or point towards q₁

Since these fields are opposite in direction.

Compute Net Field

Net Field = 1.046 x 10⁶ - 0.227  x 10⁶ N/C

               =  1046000 - 227000

               = 819000

               = 0.819 x 10⁶

               ≈ 0.82 x 10⁶

This interprets that it will act or point away from q₂

Compute force on q3

q₃ E  = 88 x 10⁻⁶ x 0.82 x 10⁶

       = 88000000  x 820000

       = 72160000000000

       = 72.16 N

Force on -ive charge in a field is always in a direction opposite to direction of field

So this interprets that direction of this field will be towards q₂.