Answer:
Explanation:
(a) For the calculation of the Electric field we use
[tex]E=\frac{V}{d}=\frac{15.0V}{1.6*10^{-3}m} =9375\frac{N}{C}[/tex]
(b) The capacitance is calculate by using the expression
[tex]C=\frac{\epsilon_{0}A}{d}=\frac{8.85*10^{-12}C^{2}/(Nm^{2})*(7.6*10^{-4}m^{2})}{1.6*10^{-3}m}=4.2*10^{-12}C[/tex]
(c) Finally, the charge on each plate is
[tex]Q=CV=(4.2*10^{-12}C)(15V)=6.3*10^{-11}C[/tex]
I hope this is useful for you
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Given Information:
Area of separation = A = 7.60 cm² = 0.000760 m²
distance = d = 1.60 mm = 0.0016 m
Potential difference = V = 15.0 V
Required Information:
(a) Electric field = E = ?
(b) Capacitance = C = ?
(b) Charge = Q = ?
Answer:
(a) Electric field = 9.375 kV/m
(b) Capacitance = 4.205 pF
(b) Charge = 63.08 pC
Explanation:
(a) The electric field between the plates kV/m
We know that electric field is given by
E = V/d
E = 15.0/0.0016
E = 9.375x10³ V/m
or
E = 9.375 kV/m
(b) the capacitance pF
The capacitance is given by
C = ε*A/d
where ε = 8.854×10⁻¹² F/m
C = 8.854×10⁻¹²*0.000760/0.0016
C = 4.205x10⁻¹² F
or
C = 4.205 pF
(c) the charge on each plate pC
The charge on each plate is given by
Q = ε*E*A
Q = 8.854×10⁻¹²*9.375x10³*0.000760
Q = 63.08×10⁻¹² C
or
Q = 63.08 pC
