Answer:
Electric field, E = 45.19 N/C
Explanation:
It is given that,
Surface charge density of first surface,
Surface charge density of second surface,
The electric field at a point between the two surfaces is given by :
E = 45.19 N/C
So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.
Answer:
The equilibrium position for the third charge is 69.28 cm
Explanation:
Given;
q₁ = -5.00 x 10⁻⁹ C
q₂ = -2.00 x 10⁻⁹ C
q₃ = 15.00 x 10⁻⁹ C
distance between q₁ and q₂ = 40.0 cm = 0.4 m
(-q₁)--------------------------------------(-q₂)---------------------------------(+q₃)
At equilibrium the repulsive force between q₁ and q₂ must be equal to attractive force between q₂ and q₃
According to Coulomb's law, repulsive or attractive force between charges is calculated as;
where;
F is repulsive or attractive force between charges
K is Coulomb's constant = 8.99 x 10⁹ Nm²/c²
r₁ is the distance between q₁ and q₂
q₁, q₂ and q₃ are the charge
distance between q₂ and q₃, r₂ is calculated as;
Therefore, the equilibrium position for the third charge is 69.28 cm