Two bidirectionally infinite line charges exist in vacuum. One has a charge density of -8 nC/m at x = -5, y = -9, while the other has a charge density of 5 nC/m at y = -7, z = -5. Determine the y-component of the electric field in V/m at (9, 5, 2). All coordinates are measured in meters.
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Two bidirectionally infinite line charges exist in vacuum. One has a charge density of -8 nC/m at x = -5, y = -9, while the other has a charge density of 5 nC/m at y = -7, z = -5. Determine the y-component of the electric field in V/m at (9, 5, 2). All coordinates are measured in meters.
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- There is a +5C charge at the center of a hollow sphere made of a perfect conductor. The hollow sphere has a net charge of +2C, an inner radius of 0.5 m and an outer radius of 0.75 m. Calculate the electric field as a function of distance from the +5C charge and the charge on each surface of the hollow sphere.A charge distribution creates the following electric field throughout all space: E(r, 0, q) = (3/r) (r hat) + 2 sin cos sin 0(theta hat) + sin cos p (phi hat). Given this electric field, calculate the charge density at location (r, 0, p) = (ab.c).A charge of Q is distributed as a quarter circle of a radius R. Lambda(theta)= lambda 0 sin theta provides the linear charge density where theta = 0 degrees along the positive x-axis. How would you find the constant of lambda in terms of Q and R, and what would the magnitude of the electric field be equal to at the center point?
- An electrical apparatus is made to test an electrical hypothesis. However, the static electric fields within and surrounding the apparatus need to be known before the test can begin. The apparatus consists of an insulating sphere with a radius of 2.50 cm placed in the center of a conducting sphere of inner radius 5.00 cm and a thickness of 1.00 cm. If the charge density of the inner insulator is ρ=4.08 C/m^3 and the total charge on the outer conducting sphere is −10.0 C, determine the theoretical electric field magnitude and direction at the following points: a) A point 0.500 cm from the center of the apparatus. b) A point 3.00 cm from the center of the apparatus. c) A point 5.75 cm from the center of the apparatus. d) A point 7.00 cm from the center of the apparatus.A cylinder of length L=5m has a radius R=2 cm and linear charge density 2=300 µC/m. Although the linear charge density is a constant through the cylinder, the charge density within the cylinder changes with r. Within the cylinder, the charge density of the cylinder varies with radius as a function p( r) =p.r/R. Here R is the radius of the cylinder and R=2 cm and p, is just a constant that you need to determine. b. Find the constant po in terms of R and 2. Then plug in values of R and 1. to find the value for the constant p. c. Assuming that L>>R, use Gauss's law to find out the electric field E inside the cylinder (rR) in terms of 1. and R. d. Based on your result from problem c, find the electric field E at r=1cm and r=4cm.The electric field along the axis of a ring-shaped charge of total charge q distributed uniformly is given by E = where R is the radius of the ring and z is the distance from the center of the ring. The electric field at the center of the ring is zero and at great distances from the ring approaches zero. At a certain distance along the z-axis the electric field strength is maximum. What is the electric field strength at this point? 9 3√√√3лE R² 9 2√√3лER² 9 5√√3лε R² 9 6√√3лE R² qz 4лεo (z² + R²) ³/2 0 q 4√3лER²
- l y d x A line charge of charge density, p = 0.0001 C/m is of length, 1=2.5 m. Find the electric field component Ę at the point, d=5 m away from one end as shown in the figure above. Answer: Ey = 4.5 × 10 V/mSuppose 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 ( ))Charge of a uniform density (11 pC/m?) is distributed over the entire xy plane. A charge of uniform density (6 pC/m2) is distributed over the parallel plane defined by z = 2.0 m. Determine the magnitude of the electric field for any point with z = 3.0 m.
- Find the electric field at the origin of the x,y-plane for charge distributions (a) and (b), see the figures shown below. The field is produced (a) by a thin half-circle with a radius of 15 cm and the linear charge density K-10 pc/cm and (b) by a thin quarter-circle with the same radius and the linear charge density K = -10 pc/cm. K>0 (a) For the charge distribution (a): The x-component of Ea. Ea,x= The y-component of Ea, Ea,y= For the charge distribution (b): The x-component of Eb, Eb,x- The y-component of Eb, Eb,y = Units N/C Units N/C Units N/C Units N/C K<0 (b)Electric Field IntegralsConsider the electric field of a point charge.At point A the electric field strength is 9 V/m, at point B the electric field strength is 36 V/m. We measure the strength of the electric eld at the midpoint of the interval AB. What can be the measured value?