The drawing shows a positive point charge +q1, a second point charge q2 that may be positive or negative, and a spot labeled P, all on the same straight line. The distance d between the two charges is the same as the distance between q1 and the spot P. With q2 present, the magnitude of the net electric field at P is twice what it is when q1 is present alone. Given that q1 = +0.20 μC, determine the agnitude |q2| when q2 is (a) positive

When a neutrally charged atom loses an electron to another atom, the result is the creation of a. two charged atoms. b. two negatively charged atoms. c. two positively charged atoms. d. two neutrally charged atoms.

Point charges q1=+2.00μc and q2=−2.00μc are placed at adjacent corners of a square for which the length of each side is 1.50 cm . point a is at the center of the square, and point b is at the empty corner closest to q2. take the electric potential to be zero at a distance far from both charges. part a what is the electric potential at point a due to q1 and q2? express your answer with the appropriate units.

An alpha particle with kinetic energy 12.5 mev makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude l = p0 b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.40×10−12 m. (assume that the lead nucleus remains stationary and that it may be treated as a point charge. the atomic number of lead is 82. the alpha particle is a helium nucleus, with atomic number 2.) what is the distance of closest approach?