Calculate the electric flux through a trapezoid, with measurements of 4 cm at the smaller base, 7 cm at the larger base and a height of 11 cm kept in the region of a uniform electric field 57 N/C (A) with the angle 0 is 60° and (B) the angle 0 zero?
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- Positive charge is distributed in a sphere of radius R that is centered at the origin. Inside the sphere, the electric field is Ē(r) = kr-1/4 f, where k is a positive constant. There is no charge outside the sphere. a) How is the charge distributed inside the sphere? In particular, find an equation for the charge density, p. b) Determine the electric field, E(r), for r > R (outside the sphere). c) What is the potential difference between the center of the sphere (r = 0) and the surface of the sphere (r = R)? d) What is the energy stored in this electric charge configuration?A non-conducting spherical shell has an inner radius & and an outer radius 28. There are no charges at r<& wherer is the distance from the center of the sphere. A total charge is distributed uniformly in the volume of the shell (between r=R and r=2R) Find the magnitude of the electric field at r=1718. Express your answer in units of using two decimal R² kQ placesThe electric field is measured for points at distances r from the center of a uniformly charged insulating sphere that has volume charge density ρ and radius R, where r<R Part A Calculate ρ.
- The charge density of a non-uniformly charged sphere of radius 1.0 m is given as: For rs 1.0 m; p(r)= Po(1-4r/3) For r> 1.0 m; p(r)= 0, where r is in meters. What is the value of r in meters for which the electric field is maximum?A quarter-disc in the first quadrant has a surface charge density given by Kcos(φ)C/m2 . If the radius of the disc is b, calculate the electric field at a point P(0, 0, −h).A positively charged cylinder has a uniform volume charge density. Height l is larger than its radius a (1»a). a P a. When Point P is very close to the surface of the cylinder (1>r>a), the electric field there can be derived by treating the cylinder as an = Eŝ. infinitely long one. Suppose that we already measure the electric field at P as charge density in terms of E and a (ŝ is the radial unit vector in the cylindrical coordinate system as defined in the Equation: now use the Gauss's law to find the volume cos o â + sin ø ŷ, - sin ø Âx + cos ø ŷ, î. b. Now we move the detector from Point P to Point Q, which is so far away from the cylinder (R>l>a), that the cylinder can be treated as a point. Based on result in Part (a), find out the electric field at Q (Note that OQ is in the x direction.)
- 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.An infinite cylindrical conductor has an inner radius ra =57.9mm and an outer radius 70.4mm. The conductor has a linear charge density of A₁ =136. On the axis of the cylinder is an infinite line charge with linear charge density ₂ = -9€. Determine the electric field magnitude at the point r = 31.03mm (in) C OThe volume charge density ρ for a spherical charge distribution of radius R= 6.00 mm is not uniform. (Figure 1) shows ρ as a function of the distance r from the center of the distribution. a)Calculate the electric field at r = 1.00 mm. b)Calculate the electric field at r = 1.00 mm.
- Consider a thin rod which has a uniformly distributed charge Qot = -1 µC. The rod is bent into a quarter of a circle of radius R = 1 m. Find the x- and y-components of the electric field created by the rod the point O the center of the arc. Hint: The following integrals are useful: cose de = [sin@]% î sine de = [-cos0]% RYou are working as an intern for a meteorological laboratory. You are out in the field taking measurements with a device that measures electric fields. You measure the electric field in the air immediately above the Earth's surface to be 139 N/C directed downward. (Assume the radius of the Earth is 6.37 x 106 m.) (a) Determine the surface charge density (in C/m²) on the ground. C/m? (b) Imagine the surface charge density is uniform over the planet. Determine the charge (in C) of the whole surface of the Earth. (e) Determine the Earth's electric potential (in V) due to the charge found in (b). V (d) Determine the difference in potential (in V) between the head and the feet of a person 1.50 m tall. (Ignore any charges in the atmosphere.) VA circular plane, with radius of 2.2 m, is immersed in an electric field with a magnitude of 800 N/C. The field makes an angle of 20° with the plane. What is the magnitude of electric flux through the plane? Calculate the electric flux passing through the surface as shown below. An infinitely long line of charge carries 0.4 C along each meter of length. Find the electric field 0.3 m from the line of charge.