Question

The plates of a parallel plate capacitor, 5.0 x 10 m apart are maintained at a potential difference of 5.0 x 10^V. Calculate the magnitude of the (i) electric field intensity between the plates. (ii) force on the electron (iii) acceleration (electronic charge of the electron. = 1.6 x 10°C mass of electron = 9.1 x 10 kg).​

Answer 1

Answer:

Given, the separation between parallel plate of capacitor is d=5.0 \times 10^{-3} \: md=5.0×10

−3

m , the potential difference is V=5.0 \times 10^4V=5.0×10

4

volt.

Now , we know that

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

⟹E=

d

V

Where, EE 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/CE=

5.0×10

−3

5.0×10

4

Newton/Columb(N/C)

⟹E=1.0×10

7

N/C

Therefore, Electric field intensity inside the capacitor is E=1.0 \times10^7 \: N/CE=1.0×10

7

N/C .