An electromagnetic wave transmits
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Answer:
Q = +1.4 pC
Explanation:
- The capacitance of any capacitor, by definition, is as follows:
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- Applying Gauss'Law and the definition of electric potential, it can be showed that the capacitance of a parallel-plate capacitor can be expressed as follows:
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- Where εr is the dielectric constant of the material that fills the space between the plates, A is the area of one of the plates, and d, is the separation between them.
- Replacing by the givens in (2) we can find the value of the capacitance C, as follows:
- Replacing the values of C and V in (1), we can solve for Q, as follows:
- As the outer surface is at a higher potential that the inside surface, the charge on it must be positive, and is equal to +1.4 pC.