Consider a closed triangular box resting within a horizontal electric field of magnitude E = 6.16 104 N/C as shown in the figure below.   A closed right triangular box with its vertical side on the left and downward slope on the right rests within a horizontal electric field vector E that points from left to right. The box has a height of 10.0 cm and a depth of 30.0 cm. The downward slope of the box makes an angle of 60 degrees with the vertical. (a) Calculate the electric flux through the vertical rectangular surface of the box. kN · m2/C (b) Calculate the electric flux through the slanted surface of the box. kN · m2/C (c) Calculate the electric flux through the entire surface of the box. kN · m2/C

Consider two infinite planes of charge.  One coincides with the x-y plane and has a charge density of +9.5 x 10-12 C/m2.  The second plane coincides with the y-z plane and has a charge density of -9.5 x 10-12 C/m2.  Calculate the magnitude of electric field, in N/C, at any location that is not on the x-y or y-z planes.        Use ε 0 = 8.9 x 10-12 F/m. (Please answer to the fourth decimal place - i.e 14.3225)

Find the x and y components of the electric field produced by q1 and q2 in the figure shown below at point A and point B.  (Take q1 = 1.88 µC and q2 = −1.16 µC.)   Point A Ex =    Ey =  Point B Ex =    Ey =

In the figure below, positive charge q = 7.47 pC is spread uniformly along a thin nonconducting rod of length L = 13.0 cm. (a) What is the magnitude of the electric field produced at point P, at distance R = 6.00 cm from the rod along its perpendicular bisector? __________N/C(b) What is the direction of the field at point P?____________ ° counterclockwise from the +x-axis   (please include units so that I can follow your steps easier)

Consider an electric field perpendicular to a table top. When a small charged object of mass 4.18 g and charge -17 µC is carefully placed in the field, the object is in static equilibrium. What are the magnitude and direction of the electric field? (Give the magnitude in N/C.) magnitude N/C direction downward

A dust particle (M=5.0x108kg) that carries charge Q=8.0x10.6℃ is placed in between two parallel plates that are equally but oppositely charged at a distance of 4.0cm from the (-) plate. The strength of this constant (uniform) electric field is E=12,500 N/C. The P.E. of Q when placed at 4.0cm from (-) plate is exactly equal to the work it can do as it is pushed to the (-) plate. is 0.004J is 400mJ Answeres a, b & c are correct. Answeres a & b are correct.

could you solve this

Two identical thin rods of length L are each charged with a positive charge q that is uniformly distributed on each rod. The rods are arranged parallel to each other along the x direction and are separated by a distance 2d, as shown in the figure below. Determine the electric field at point P, which is located a distance d from the right endpoint of each rod. Hint: You will eventually need the following integral: น 1 S (a²+u²)3/2 du = − + C √a²+u²

The point charges in the figure below are located at the corners of an equilateral triangle 20.0 cm on a side. (Assume that the +x-axis is to the right and the +y-axis is up along the page.) (a) Find the electric field (in N/C) at the location of  qa in the figure, given that qb = +19.00 µC and qc = −8.00 µC.   magnitude _______ N/C   direction __________° counterclockwise from the +x-axis   (b) What is the force (in N) on qa, given that qa = +2.50 nC?   magnitude ______N   direction _________° counterclockwise from the +x-axis