Answer: a) 14*10^3 V; b)2.4 *10^-3 m^2; c)6.07*10^-12 F
Explanation: In order to explain this problem we have to consider, the following expressions:
V=E*d the difference of potencial in a capacitor is equal to the electric field multiply the separacion between the plates then,
V= 4*10^6*3.5*10^-3=14*10^3 V
We also know that the electric field in a parallel plates capacitor is given by:
E=σ/εo where σ is Q/A ( charge/area)
E=Q/(A*εo)=85*10^-9/(A*8.85*10^-12) then
A=85*10^-9/(8.85*10^-12*4*10^6)=2.4 *10^-3 m^2
Finally, the capacitance is given by:
C= εo*A/d
C=8.85*10^-12*2.4*10^-3/3.5*10^-3=6.07*10^-12 F
The capacitance of the capacitor is 6.07 * 10^-12 F.
a) To obtain the potential difference between the plates;
V=E*d
Where;
E = magnitude of the electric field
d = distance between the plates
Hence;
V= 4*10^6 * 3.5*10^-3= 14 * 10^3 V
b)We also know that so we can get the area from;
E=σ/εo and σ = Q/A hence
E=Q/(A*εo)=85*10^-9/(A*8.85*10^-12) therefore
A=85*10^-9/(8.85*10^-12*4*10^6)
=2.4 * 10^-3 m^2
c) For the capacitance is given by:
C= εo*A/d
C=8.85*10^-12*2.4*10^-3/3.5*10^-3
=6.07 * 10^-12 F
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