Answer:
The magnitude of the electric field is 1.124 X 10⁷ N/C
Explanation:
Magnitude of electric field is given as;
where;
E is the magnitude of the electric field, N/C
q is the point charge, C
k is coulomb's constant, Nm²/C²
r is the distance of the point charge, m
Given;
q = 5mC = 5×10⁻³ C
r = 2m
k = 8.99 × 10⁹ Nm²/C²
Substitute these values and solve for magnitude of electric field E
Therefore, the magnitude of the electric field is 1.124 X 10⁷ N/C
Answer:
so pressure in A must be one third the pressure in B
Explanation:
We shall apply gas law to the cylinders A and B . Since their quantity are same so their no of mole will also be same .
For cylinder A
Temperature T , volume 3V , pressure P₁ , no of mole = n
so
P₁ X 3V = n R T
For cylinder B
Temperature T , volume V , pressure P₂ ,no of mole = n
so
P₂ X V = n R T
From the two equation above
P₁ X 3V = P₂ X V
P₁ = P₂ / 3
so pressure in A must be one third the pressure in B