22.15 (a) A conducting sphere has charge Q and radius R. If the electric field of the sphere at a distance r = 2R from the center of the sphere is 1400N/C, what is the electric field of the sphere at r = 4R? (b) A very long conducting cylinder of radius R has charge per unit length I. Let r be the perpendicular distance from the axis of the cylinder. If the electric field of the cylinder at r = 2R is 1400N/C, what is its electric field at r = 4R? (c) A very large uniform sheet has a surface charge density s. If the electric field of the sheet has a value of 1400N/C at a perpendicular distance d from the sheet, what is the electric field of the sheet at a distance 2d from the sheet?

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).

An infinitely long cylinder of radius R = 57 cm carries a uniform charge density ρ = 21 μC/m 3. Calculate the electric field (in N/C) at distance r = 19 cm from the axis of the cylinder.

A square conducting plate 54.0 cm on a side and with no net charge is placed in a region, where there is a uniform electric field of 75.0 kN/C directed to the right and perpendicular to the plate. (a)Find the charge density (in nC/m2) on the surface of the right face of the plate. (b)Find the charge density (in nC/m2) on the surface of the left face of the plate. (c)Find the magnitude (in nC) of the charge on either face of the plate.

A uniform electric field of magnitude 5.5 x 104 N/C passes through the plane of a square sheet with sides 9.0 m long. Calculate the flux (in Nm²/C) through the sheet if the plane of the sheet is at an angle of 30° to the field. Find the flux for both directions of the unit normal to the sheet. unit normal with component parallel to electric field Nm²/C unit normal with component antiparallel to electric field Nm²/c Additional Materials

A uniform electric field of magnitude 3.6×104N/C is at an angle of 80° to a square sheet with sides 8.5 m long. What is the electric flux through the sheet?

An electric field 7 N/C is passing through a region of area 2.1 m² in the xy-plane. If the electric field makes an angle of 56° with respect to the area vector perpendicular to the region, calculate the electric flux magnitude (in N-m²/C) through the region.

A vertical wall (6.3 mx 3.3 m) in a house faces due east. A uniform electric field has a magnitude of 210 N/C. This field is parallel to the ground and points 34° north of east. What is the electric flux through the wall? Number: Units

For a given surface, the electric flux, PE, is proportional to the number of field lines through the surface. For a uniform electric field, the maximum electric flux is equal to the product of electric field at the surface and the surface area (i.e., EA). The electric flux is defined to be positive when the electric field E has a component in the same direction as the area vector A and is negative when the electric field has a component in the direction opposite to the area vector. C. Sketch vectors A and Ē such that the electric flux is: Positive Negative Zero