Answer to Question #115419 in Electricity and Magnetism for Jafar Abbas

Question

The plates of a parallel plate capacitor,5.0x10^-3m apart are maintained at a potential difference of 5.0x10^4 Calculate the magnitude of the I. Electric field intensity between the plates II. Force on the electron.

Answer 1

Given, the separation between parallel plate of capacitor is $d=5.0 \times 10^{-3} \: m$ , the potential difference is $V=5.0 \times 10^4$ volt.

Now , we know that

$$V=E\cdot d \\\implies E=\frac{V}{d}$$

Where, $E$ denotes the electric field intensity inside the capacitor.

i). Now, from the above formula we get,

$$E=\frac{5.0 \times 10^4}{5.0 \times 10^{-3}} \: \text{Newton/Columb}(N/C)\\ \implies E=1.0 \times10^7 \: N/C$$

Therefore, Electric field intensity inside the capacitor is $E=1.0 \times10^7 \: N/C$ .

ii). Since, force acting on a charge due to electric field is given by

$$F=e \cdot E$$

and S.I unit is Newton $(N)$, where $e = 1.602 \times 10^{-19} C$ is the magnitude of the charge on a electron.

Hence, force acting on the electron is,

$$F=1.602 \times 10^{-19} \cdot( 1.0 \times 10^7)\\ \implies F=1.602\times10^{-12} \: N$$

in the opposite direction of electric field.

Hence, we are done.