Parallel Plate Capacitors

A uniform electric field is established by connecting the plates of a parallel-plate capacitor to a 12-V battery. A) If the plates are seperated by 0.75 cm, what is the magnitude of the electric field in the capacitor? B) A charge of +6.24 x 10 -6 C moves from the positive plate to the negative plate. Find the change in electric potential energy for this charge.

The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. The amount of charge on the plates is now equal to (a) Q/2. (b) Q. (c) Q/4. (d) 2Q. (e) 4Q.

Two parallel-plate capacitors C1 and C2, are separated by vacuum. C1 plate has an area twice as large as C2, but the plates of C1 are separated by a distance 1/3 that of C2. What is the ratio of their capacitance (C1: C2)? A) 1:6 B) 6:1 C) 2:3 D) 3:2

Consider a parallel-plate capacitor constructed from two circular metal plates of radius R. The plates are separated by a distance of 1.5 mm. (a) What radius must the plates have if the capacitance of this capacitor is to be 1.0 μF? (b) When the capacitor is connected to a 12.0-V battery, what is the magnitude of the charge on each of the plates? (c) Find the energy stored in the capacitor.

A parallel plate capacitor has plates with area A = 350cm2 separated by a distance d = 1.5mm. What is the capacitance when the capacitor is filled with air?  

You reposition the two plates of a parallel-plate capacitor so that the distance between them doubles. There is vacuum between the plates. If the charges +Q and -Q on the two plates are kept constant in this process, which of the following statements is incorrect? A) The potential difference between the two plates doubles. B) The electric field between the two plates keeps the same. C) The potential energy stored in this capacitor doubles. D) The capacitance of this capacitor doubles. 

Each plate of a parallel-plate air capacitor has an area of 0.0010 m 2, and the separation of the plates is 0.050 mm. An electric field of 7. 4x10   6 V/m is present between the plates. The capacitance of the capacitor, in pF, is closest to: A) 180 B) 300 C) 240 D) 120 E) 360

The plates of a parallel-plate capacitor are 3.0mm apart, and each plate carries 75nC of charge. The electric field between the plates has a magnitude of 3.5 x 106V/m. The plates are in vacuum. a) What is the potential difference between the plates? b) What is the capacitance? c) What is the area of each plate?

The potential difference between the plates of a parallel plate capacitor with the plate separation of 6 mm is 60 V. What is the electric field between the plates of this capacitor?A) 500 V/mB) 1000 V/mC) 2000 V/mD) 60 V/m

What property of objects is best measured by their capacitance?a) ability to conduct electric currentb) ability to distort an external electrostatic fieldc) ability to store charge

Consider an air-filled charged parallel plate capacitor. How can its capacitance be increased?a) Increase the charge on the capacitor.b) Decrease the charge on the capacitor.c) Increase the spacing between the plates of the capacitor.d) Decrease the spacing between the plates of the capacitor.e) Increase the length of the wires leading to the capacitor plates.

(a) Assume that charge −q is placed on the top plate, and +q is placed on the bottom plate. What is the magnitude of the electric field E between the plates? Express E in terms of q and other quantities given in the image, in addition to ε0 and any other constants needed.(b) What is the voltage V between the plates of the capacitor? Express V in terms of the quantities given in the image and any required physical constants.(c) Now find the capacitance C of the parallel-plate capacitor. Express C in terms of quantities given in the image and constants like ε0.

Consider a charged parallel-plate capacitor. Which combination of changes would quadruple its capacitance?a) Double the charge and double the plate area.b) Double the charge and double the plate separation.c) Halve the charge and double the plate separation.d) Halve the charge and double the plate area.e) Halve the plate separation; double the plate area.f) Double the plate separation; halve the plate area.