what is b plz . Consider a disconnected parallel plate capacitor with area Ao and width --4, 3do. There is a magnitude of charge density g, on each plate. For this problem you may assume the plates ar...
A parallel-plate capacitor has plates of area 0.13 m2 and a separation of 1.9 cm. A battery charges the plates to a potential difference of 150 V and is then disconnected. A dielectric slab of thickness 9.3 mm and dielectric constant 4.0 is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted? (b) What is the capacitance with the slab in place? What is the free charge q (c) before and (d) after the slab is inserted? What is the magnitude of the...
A parallel plate capacitor is made of plates of area 0.05 m? each. The plates are separated by a distance of 0.200 mm. Initially, the space between the plates is filled with air. (a) What is the capacitance of this air-filled capacitor? (b) If the electric field inside the capacitor exceeds 3.00 x 106 V/m, the air undergoes electrical break- down. (This maximum field is known as the dielectric strength of air.) From this, calculate the maxi- mum voltage (potential...
The plates of a parallel-plate capacitor are 700 cm2 in area and 0.3 cm apart. The potential difference between the plates is 935 V. (a) What is the magnitude of the electric field between the plates? (b) the charge on each plate? (c) the force exerted by the field on one of the plates? Now the two plates touch at the bottom forming a V-shape. (d) Find the total electric energy stored between the plates.
003 (part 1 of 2) 10.0 points Consider an air-filled parallel plate capacitor with plate area A and gap width d and plate charge What is the total energy stored in the ca pacitor? 2 0 A Eo d Eo A d 2 0 A d 4. U = 004 (part 2 of 2) 10.0 points With the battery connected, fill the gap by a slab with the dielectric constant 3.6. If the potential is 30 V, the plate sep-...
1) The plates of a parallel-plate capacitor are 700 cm2 in area and 0.3 cm apart. The potential difference between the plates is 935 V. (a) What is the magnitude of the electric field between the plates? (b) the charge on each plate? (c) the force exerted by the field on one of the plates? Now the two plates touch at the bottom forming a V-shape. (d) Find the total electric energy stored between the plates.
A parallel-plate capacitor (Homework #2) A parallel-plate capacitor is made by a pair of plates with area A and separation d. At first, the plates are charged up to +Q and -O respectively, and then a battery is disconnected. Express all the answers for a) through f) only by Q, d, A and 5 +Q 2d -Q Area A a) What is the electric field E inside of the capacitor? b) What is the potential difference V between the plates?...
explain why please
A parallel-plate capacitor is made of two conducting plates of area A separated by a distance d. The capacitor carries a charge Q and is initially connected to a battery that maintains a constant potential difference between the plates. The battery is then disconnected from the plates and the separation between the plates is doubled. ) Which of the following remains constant? Voltage across the capacitor Capacitance of the capacitor Charge on the capacitor Submit (Survey Question)...
A parallel-plate capacitor has plate area of 0.12 m2and plate separation 1.2 cm. It is charged by a battery to potential difference of 120 V, then disconnected. A dielectric slab, thickness 4.0 mm and dielectric constant 4.8, is placed symmetrically between the plates.(a) Calculate the capacitance before and after the slab is inserted.(b) Calculate the free charge q before and after the slab is inserted.(c) Calculate the magnitude of the electric field in the space between the plates and the...
2) The plates of a parallel-plate capacitor are 700cm2 in area and 0.3 cm apart. The potential difference between the plates is 935 V. (a) What is the magnitude of the electric field between the plates? (b) the charge on each plate? (c) the force exerted by the field on one of the plates? Now the two plates touch at the bot- tom forming a V-shape. (d) Find the total electric energy stored between the plates. In d) is U=0J?...
There is a parallel plate capacitor with a separation of 0.020 m and an area of 0.40 m^2. The charge magnitude is 3.6x10^-6 C. What is the electric field? What would the electric potential difference (Va-Vb) be between the (point a) negative plate and a point halfway between (point b)? If an electron is moved between points a and b how much does the electron's potential energy change? Also how much work is done?