A parallel-plate capacitor with area 0.600m 2 and plate separation of 2.00mm is

Question

A parallel-plate capacitor with area 0.600m2 and plate separation of 2.00mm is connected to a 21.00 V battery. (a) What is the capacitance?
1 F

(b) How much charge is stored on the plates?
2 C

(c) What is the electric field between the plates?
3 N/C

(d) Find the magnitude of the charge density on each plate.
4 C/m2

(e) Without disconnecting the battery, the plates are moved fartherapart. Qualitatively, what happens to each of the previousanswers?

5 (a) What is the capacitance?
1 F

(b) How much charge is stored on the plates?
2 C

(c) What is the electric field between the plates?
3 N/C

(d) Find the magnitude of the charge density on each plate.
4 C/m2

(e) Without disconnecting the battery, the plates are moved fartherapart. Qualitatively, what happens to each of the previousanswers?

5 5

Explanation / Answer

Given that A parallel-plate capacitor with area (A) =0.600 m2 and plate separation of (d) = 2.00 mm The paltes are connected to a battery (V)= 21.00 V The permittivity (0)=8.854*10-12F/m a) Then the capacitance is                       C =0A/d b) The charge is stored on the plates is                      Q =CV       ( C calculated from thepart (a) ) c) The electric field between the plates is                       E=V/d d) The magnitude of the charge density on each plate.is                    = Q/A         (Q calculated from the part (b) ) e) Capacitance : Decreases Charge : Decreases Electric field : Decreases Charge density : It does not depends on the distance (So, it remains same ) Now substitute all the above values you get the requiredanswer

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