The figure shows four (4) closed surfaces which enclose electrical charges. If the value of Q = 2.25 x flux on surfaces S 1 and S2. 10-10 C, determine the electrical -2Q +2 S₂
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- - C. Problems C1. For the configuration of conductors depicted below, find the electrodynamic force F2 = ? developed in conductor 2, given: 3 I =1000 A L=2m I, I I2 = 2500 A а%3D2 ст I3 = 3000 A b= 6 cm L I4 = 5000 A c = 4 cm bA volume charge distribution in free space has p,=1.8r (nC/m³) for 0srs5m (inside of sphere of radius r), 0s8spi, 0s0s2pi, and zero otherwise. Determine magnitude of Electrical field (in V/m) at r=1m. Permittivity of free space is given as ɛ,=(1/(36pi)x10-° F/m. O a. 61.07 O b. 50.89 О с. 55.98 O d. 43.26 О е. 35.63Problem 6: The voltage across a membrane forming a cell wall is 83 mV and the membrane is 9.25 nm thick. What is the electric field strength in the cell wall, in volts per meter? E = sin() cotan() atan() acotan() cosh() tanh() cos() asin() O Degrees tan() π ( ) 789 acos() E ^^^ 4 5 6 sinh() 1 1 2 3 cotanh() + 0 Radians VOBACKSPACE HOME END DEL CLEAR
- Under electrostatic conditions, the excess charge on a conductor resides on its surface. Does this mean that all the conduction electrons in a conductor are on the surface?Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rp has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = aer/ao + B/r + bo %| where alpha (a), beta (B), ao and bo are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Va Edr= Edr Calculating the antiderivative or indefinite integral, Vab = (-aaoe-r/ao + B + bo By definition, the capacitance C is related to the charge and potential difference by: C = Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q/( (e-"b/ao - era/ao) + B In( ) + bo ( ))A block in the shape of a rectangular solid has a cross-sectional area of 4.00 cm^2 across its width, a front-to-rear length of 5.00 cm, and a material conductivity of 10^6 Ω x m^-1. The block’s material contains 5.00x10^22 conduction electrons/m^3. A potential difference of 35.0 V is maintained between its front and rear faces. Find: (a) the resistivity of the block material(b) the resistance of the block(c) the current through the block(d) the magnitude of the current density if it is uniform(e) the drift velocity of the conduction electrons (f) the magnitude of the electric field through the block
- A ring-shaped conductor with radius a = 2.70 cm has a total positive charge Q = 0.125 nC uniformly distributed around it. (Figure 1) For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Field of a ring of charge. Figure dQ ds X dE dE α 1 of 1 dE X SET UP: Q = 0.125 × 10-⁹ C, a = 0.027 m and x = 0.430 m. 1 Qx EXECUTE: E Απεο (22+a2)3/2° Part B +x-direction -x-direction What is the direction of the electric field at point P? Submit Previous Answers Correct Part C ·î · = |VD | ΑΣΦ F = 1.7910-5 (6.04 N/C)i. A particle with a charge of - 2.40 μC is placed at the point P described in part A. What is the magnitude of the force exerted by the particle on the ring? Express your answer in newtons. Submit Previous Answers Request Answer ? Review | Constants NThe plane z = 0 marks the boundary between free space (z 0 side) with a relative permittivity of e, =35. The electric field intensity next to the interface in free space is Ē = –10£ + 25ŷ + 92 V/m. Determine the electric field intensity on the other side of the interface.z Q. The ends of the cylinder are made of conducting discs and the space between these discs are filled with an inhomogeneous ohmic medium whose conductivity o = L/z'. Where L is the separation distance of the L discs. A d-c voltage Vo is applied across the discs as shown on the right. Determine V. a) The total resistance between the discs b) The surface charge densities on the discs c) The volume charge density and the total amount of charge between the discs y N
- It has been experimentally observed that the electric field in a large region of earth's atmosphere is directed verti- cally down. At an altitude of 300 m, the electric field is 60 Vm-¹. At an altitude of 200 m, the field is 100 Vm¯¹. Calculate the net amount of charge contained in the cube of 100 m edge, located between 200 and 300 m altitude. E₁ = 60 Vm-1 E₂ = 100 Vm-1 Fig. ds E dS K 300 m Ť 200 mIn an electrically neutral insulator, electrons are *not* able to move around freely as they can in a conductor. How, then, is it possible for an external electric field to induce a charge distribution in a chunk of insulating material?A wire with 4 meters length and constant charge density lambda = 4nC/m is placed diagonally on x-y plane. One end is at the origin and it makes 45 degrees with the x-axis. Find the integral expression for Ex at x=7 y=10 (Please help as soon as available. Thank you very much in advance)