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

Answer:

A. $E = 1.512\times 10^{5}\ V$

B. 151.2 V

C. $E = 1.512\times 10^{5}\ V$

D. V = 302.4 V

Solution:

As per the question:

Area of the plates of the parallel plate capacitors, A = $2.30\times 2.30 = 5.29\ cm^{2} = 5.29\times 10^{- 4}\ m^{2}$

Charge on the plates of the capacitor, $Q_{c} = \pm 0.708\ nC = \pm 0.708\times 10^{- 9} \C$

Now,

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

$E = \frac{Q}{\epsilon_{o}A}$

$E = \frac{0.708\times 10^{- 9}}{8.85\times 10^{- 12}\times 5.29\times 10^{- 4}} = 1.512\times 10^{5}\ N/C$

(B) To calculate potential difference between the plates:

$V = Ed = 1.512\times 10^{5}\times 10^{- 3} = 151.2V$

(C) Electric field strength when spacing is 2 mm, i.e., $2\times 10^{- 3}\ m$:

$E = \frac{Q}{\epsilon_{o}A}$

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., $1.512\times 10^{5}\ V$

(D) Potential difference across the capacitor when d = $2\times 10^{- 3}\ m$:

V = Ed = $1.512\times 10^{5}\times 2\times 10^{- 3} =302.4\ V$