Imagine that two parallel plates are charged to achieve a voltage difference Vo between the plates. They have area A and spacing d. The plates are then electrically isolated so that no charge can be added or taken away. A good conductor of thickness d/2 is placed in between the two plates, as shown. Given that the charge on the plates cannot change, what is the electric field in the regions between the plates, but above and below the conductor? (If you are stuck, Gauss’s Law may help.)

Consider a thin plastic rod bent into an arc of radius R and angle a. The rod carries a uniformly distributed negative charge -Q. (Note: the diagram may have the incorrect sign.) y α R -Q X What is the x component of the electric field at the origin? (Enter your responses in terms of the symbolic quantities mentioned in the problem. To make things easier, just write the letter "a" for the angle a, and use the Coulomb constant k rather than the unwieldy 1/4no.) Ex = k*Q*sin(a)/(a*R^2) Computer's answer now shown above. Tries 0/6 What is the y component of the electric field at the origin? Ey = k*Q*(1-cos(a))/(a*R^2) Computer's answer now shown above. Tries 0/6 Follow the steps outlined in class and in the textbook: 1. Use a diagram to explain how you'll cut up the charge distribution, and draw the AE contributed by a representative piece of charge at a given location. 2. Express algebraically the contribution each piece makes to each vector component of the electric field. Indicate…

We have two uniformly charged parallel plates, as shown. Their widths are much larger than the separation between the plates. The magnitudes of the charges on each are equal. a, b, and c indicate the regions just above, in between, and just below the plates, respectively. Now let's suppose that the plates are squares with side length 1 cm, and that they have equal and opposite charges of magnitude 3.7 μC. What is the magnitude of the electric field in region b?

A small block of mass m and charge Q is placed on an insulated, frictionless, inclined plane of angle as in the figure below. An electric field is applied parallel to the incline. Q m E = (a) Find an expression for the magnitude of the electric field that enables the block to remain at rest. (Use any variable or symbol stated above along with the following as necessary: g for the acceleration due gravity. Note that the charge of Q is unknown.) i (b) If m = 5.41 g, Q = -7.49 μC, and 0 = 26.3°, determine the magnitude and the direction of the electric field that enables the block to remain at rest on the incline. magnitude N/C direction ---Select---