Answer to Question #107363 in Optics for Mukelo Madolo

Question

an airfilled parallelplate capacitor has plates of area 2.30cm^2 seperated by 1.50 mm. a)find the value of its capacitance . the capacitor is connected to a 120 v battery. b)what is the charge on the capacitor c)what is the magnitude of the uniform electric field between the plates?

Answer 1

(a) The capacitance

$$C=\frac{\epsilon_0A}{d}=\frac{8.85\times 10^{-12}\times 2.30\times 10^{-4}}{1.50\times 10^{-3}}=1.36\times 10^{-12}\:\rm F$$

(b) The charge on plates

$$q=CV=1.36\times 10^{-12}\times 120=1.63\times 10^{-10}\:\rm C.$$

(c) The electric field between plates

$$E=\frac{V}{d}=\frac{120}{1.50\times 10^{-3}}=80\: 000\:\rm V/m.$$