Question

Two 2.30 cm × 2.30 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC . Part A What is the electric field strength inside the capacitor if the spacing between the plates is 1.00 mm ? Express your answer with the appropriate units. E E = nothing nothing SubmitRequest Answer Part B What is potential difference across the capacitor if the spacing between the plates is 1.00 mm ? Express your answer with the appropriate units. V V = nothing nothing SubmitRequest Answer Part C What is the electric field strength inside the capacitor if the spacing between the plates is 2.00 mm ? Express your answer with the appropriate units. E E = nothing nothing SubmitRequest Answer Part D What is the potential difference across the capacitor if the spacing between the plates is 2.00 mm ? Express your answer with the appropriate units. V V = nothing nothing SubmitRequest Answer Provide Feedback Next

Answer 1

A.
`E = 1.512* 10^(5)\ V`

B. 151.2 V

C.
`E = 1.512* 10^(5)\ V`

D. V = 302.4 V

Solution:

As per the question:

Area of the plates of the parallel plate capacitors, A =
`2.30* 2.30 = 5.29\ cm^(2) = 5.29* 10^(- 4)\ m^(2)`

Charge on the plates of the capacitor,
`Q_(c) = \pm 0.708\ nC = \pm 0.708* 10^(- 9) \C`

Now,

(A) To calculate the electric field strength, E when the separation distance, d = 1.00 mm =
`10^(- 3)\ m`:

`E = (Q)/(\epsilon_(o)A)`

`E = (0.708* 10^(- 9))/(8.85* 10^(- 12)* 5.29* 10^(- 4)) = 1.512* 10^(5)\ N/C`

(B) To calculate potential difference between the plates:

`V = Ed = 1.512* 10^(5)* 10^(- 3) = 151.2V`

(C) Electric field strength when spacing is 2 mm, i.e.,
`2* 10^(- 3)\ m`:

`E = (Q)/(\epsilon_(o)A)`

Since, the above expression of the electric field shows that it does not depend on the separation distance between the plates thus it will remain same, i.e.,
`1.512* 10^(5)\ V`

(D) Potential difference across the capacitor when d =
`2* 10^(- 3)\ m`:

V = Ed =
`1.512* 10^(5)* 2* 10^(- 3) =302.4\ V`