Find the first non-zero multipole expansion term for voltage far away from a sphere of radius R and surface charge o (0) = 0, cos cos 0. %3D
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- Solve e. Use this if necessary: The dipole moment of the given system is-- x component= 18.3*10^-16 C.m y component= -122.61*10^-16 C.mIn dielectric material with ε = 2.8 ε0 , V = sin(3φ). At point P (1, 5°, 4), the volumetric charge density equals to?ato tuonogr→ 1. 'rl Use the boundary condition to find E2 in the medium 2 with boundary located at plane y = 0. Medium one is perfect dielectric characterized by &₁₁ = 3, medium 2 is perfect dielectric characterized by &₁2 = 5, electric field in medium 1 is E₁ = 3âx + 2ây + â₂ V/m. 6=0 gmon 13 y>0 obrisod mort E2 Er2 = 5 y > 607 Answer: E2= 3âx +1.2ây + âz V/m 02: 3.3 109 y = 0
- A charge ±Q is placed on the plates of a capacitor, which is then disconnected from the outside world. The separation between the plates is changed, and it is found that the voltage across the plates decreases. (a) Have the plates been moved closer together or farther apart? O closer together O farther apart (b) If the new voltage is smaller than the original voltage by a factor of six, what is the ratio of the new plate spacing to the original spacing?A parallel-plate capacitor with plate separation d is connected to a battery that has potential difference AVbat. Without breaking any of the connections, insulating handles are used to increase the plate separation to 2d.4) (a)Using the Gauss law find the electric field of an infinite conductor planewith charge density σ.(b)Using the result in part (a) calculate the capacitance of a capacitor made oftwo conducting plane of 2 m2 area and with charge density σ = 9.6 C/m 2 andthese conducting planes are separated with a distance 0.1 mm and there isAluminium oxide as the dielectric between two plates. *Take gravitational constant g =10 m/s2.
- 2 / 2 150% 4. A potential difference V, is applied, with a battery, to the plates of a parallel-plate capacitor of plate-area A and separation d. Write an expression for each of the following (a) the capacitance C, of the capacitor, (b) the free charge qo, (c) the electric field strength E, in the gap. Now the battery is disconnected and a dieleetric slab of thickness b and dielectric constant k is inserted in the gap. (d) What happens to the free charge? (c) What happens to the field strength in the gap? () What is the field strength in the dielectric slab? (g) Find an expression for the potential difference between the two plates. (h) Find an expression for the capacitance C now. (i) Assume that C becomes double of C, when a slab of dielectric constant 10 is nserted. What is the thickness b? AiCan you solve the question step by step clearly ?When can the approximation of infinite of large plates be applied? O it can always be approximated When the square root of the area of the plates is much greater than the separation distance When the square root of the separation distance of the plates is much greater than the surface area O When the capacitor is very small