Complete Question
A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148μC on each plate. The plates are 0.328 mm apart.What is the magnitude of the surface charge density on each plate?
Answer:
The value is [tex]\sigma = 1.63 *10^{-7} C/m^2 [/tex]
Explanation:
Generally the surface charge density is mathematically represented as
[tex]\sigma = E * \epsilon_o[/tex]
Here E is the electric field which is mathematically evaluated as
[tex]E = \frac{Q }{C* d }[/tex]
So substituting [tex]245 pF = 245 *10^{-12}\ F[/tex] for C , [tex]0.328\ mm = 0.328*10^{-3} \ m[/tex] for d and [tex]0.148\mu C = 0.148*10^{-6} \ C[/tex] for Q we have
[tex]E = \frac{0.148*10^{-6} }{245 *10^{-12}* 0.328*10^{-3} }[/tex]
[tex]E = 1840 *10^{3} \ V[/tex]
Hence
[tex]\sigma = 1840 *10^{3} * 8.85*10^{-12}[/tex]
[tex]\sigma = 1.63 *10^{-7} C/m^2 [/tex]
