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.
