Answer:
Part A: 7500 V
Part B: 2.899×10⁻³ m²
Part C: 10.27 pF or 10.27×10⁻¹² F
Explanation:
Part A:
Applying,
E = V/d................ Equation 1
Where E = electric field intensity between the plates, V = potential difference between the plates, d = distance of separation between the plates
make V the subject of the equation above,
V = Ed............. Equation 2
Given: E = 3.0×10⁶ V/m, d = 2.5 mm = 2.5×10⁻³ m
Substitute into equation 2
V = 3.0×10⁶ (2.5×10⁻³ )
V = 7.5×10³ V
V = 7500 V
Part B:
Using,
E = Q/(e₀A).................... Equation 3
Where Q = Charge on each plate of the capacitor, A = Area of each plate, e₀ = constant = dielectric = permitivity of free space
make A the subject of the equation,
A = Q/(e₀E).............. Equation 4
Given: Q = 77 nC = 77×10⁻⁹ C, E = 3.0×10⁶ V/m
Constant: e₀ = 8.854×10⁻¹² F/m
Substitute into equation 4
A = 77×10⁻⁹/(8.854×10⁻¹²× 3.0×10⁶)
A = 77×10⁻⁹/(26.562×10⁻⁶)
A = 2.899×10⁻³ m²
A = 2.899×10⁻³ m².
Part C:
Using,
Q = CV.................. Equation 5
Where C = Capacitance of the capacitor
make C the subject of the equation
C = Q/V.............. Equation 6
Given: Q = 77 nC = 77×10⁻⁹ C, V = 7500 V
Substitute into equation 6
C = 77×10⁻⁹/7500
C = 10.27×10⁻¹² F
C = 10.27 pF
