Two charged, square plates, separated by a distance of 22.8 cm and each with a side length of 37.4 cm, make up a parallel-plate capacitor. The electric field inside the plates is 2.00×103 N/C. If the charge is doubled, what would the electric field inside the plates become?
Two charged, square plates, separated by a distance of 22.8 cm and each with a side...
Question 3 Two charged, square plates, separated by a distance of 15.9 cm and each with a side length of 21.5 cm, make up a parallel-plate capacitor. The electric field inside the plates is 1.00x103 N/C If the charge is doubled, what would the electric field inside the plates become? 1.00E+03 N/C 250. N/C O 2.00E+03 N/C O 500. N/C
Question 2 A charged sphere has a radius of 26.9 cm. The electric field strength at a distance 34.0 cm is 1.20x103 N/C. If the sphere's charge is halved, what would the strength of the electric field at 34.0 cm become? 2.40E+03 N/C 600. N/C 1.20E+03 N/C 300. N/C Question 3 Two charged, square plates, separated by a distance of 5.40 cm and each with a side length of 18.1 cm, make up a parallel-plate capacitor. The electric field inside...
Each plate of a parallel‑plate capacitor is a square of side 0.0309 m, and the plates are separated by 0.575×10−3 m. The capacitor is charged and stores 8.77×10−9 J of energy. Find the electric field strength ?E inside the capacitor. in N/C
A parallel plate capacitor has square plates with sides of length 11 cm. The distance between the plates is 2 mm. The plates are charged up to 20volts. Part A What is the electric field between the plates? Express your answer using three significant figures. Electric field = N/C Part B What is the amount of charge on each plate? charge = C Part C What is the capacitance? Capacitance = μF Part D What is the energy stored by...
Each plate of a parallel-plate capacitor is a square of side 0.0387 m, and the plates are separated by 0.503 x 10-3 m. The capacitor is charged and stores 7.57 x 10-9 J of energy. Find the electric field strength E inside the capacitor. E-9.37 ×10 N/C
A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±Q. Part A What is the ratio Ef/Ei of the final electric field strength Ef to the initial electric field strength Ei if Qis doubled? EfEi= Part B What is the ratio Ef/Ei of the final electric field strength Ef to the initial electric field strength Ei if Lis doubled? Part C What is the ratio Ef/Ei of the final...
Two square metal plates are placed parallel to each other, separated by a distance d= 2.28 cm. The plates have sides of length L = 0.690 m. One of the plates has charge Q= + 2.34×10-6 C, while the other plate has charge -Q. Calculate the magnitude of the electric field between the plates, not close to the edge, i.e., assume a uniform surface charge distribution.
Square parallel conducting plates each have side lengths of 90 cm and are separated by distance of 3 mm. The plate on the left carries uniformly distributed surface charge of 144 nC. The side view is shown in the figure, which is not to scale. Calculate the magnitude abd direction of the pressure experienced by the plate on the left. Pressure is defined as force per unit of area.
A parallel plate capacitor is formed with two plates separated by 5.00 mm as shown in Figure 1. Each plate is a 10.0 cm X 10.0 cm square. We do not know the charge on the plates. An electron beam is shot in from one edge of the capacitor. It enters the capacitor very close (call it 0 mm) from the top plate and travelling parallel to the plates. The electrons in the beam are moving at 1.50 × 107...
A parallel plate capacitor is formed with two plates separated by 5.00 mm as shown in Figure 1. Each plate is a 10.0 cm X 10.0 cm square. We do not know the charge on the plates. An electron beam is shot in from one edge of the capacitor. It enters the capacitor very close (call it 0 mm) from the top plate and travelling parallel to the plates. The electrons in the beam are moving at 1.50 × 107...