Problem 5: Two point charges totaling 6.5 µC exert a repulsive force of 0.055 N on one another when separated by 0.45 m. Randomized Variables Q = 6.5 µC F = 0.055 N r= 0.45 m Part (a) What is the charge on the larger of the two charges in coulombs? 91 = Part (b) What is the charge on the smaller of the two charges in coulombs? (Note: If you calculate your answer using your result from part (a), make sure you didn't round it substantially.) 92 = Part (c) What is the charge in coulombs on the larger charge (in magnitude) if the force is attractive? g'1 =

The diagram below shows two point charges, A & B. The charge of A is 1.9 nC & the charge of B is -7.1 nC. Determine the magnitude & direction of the net E-field at point P. The distance between A & P is 0.31 m & the distance between B & P is 0.41 m. P WebAssign Plot P ●? 14 cm- Part B: Location of zero E-field Two point charges, A & B (qA = -20 μC & qB = 50 μC), are separated by 14 cm. The electric field at point P (see below) is zero. How far from A is P? A Enet = direction of Enet = B o relative to the HORIZONTAL in quadrant 3

Three charges are located on an x-y axis.  q1 has a value of 2.0 mC and is at (0, 3m).  q2 has a value of 3.0 mC  and is located at the origin.  q3 has a value of 4.0 mC and is located at (4 m, 0).  Find the magnitude and direction of the force on q1.  Don't forget to convert mC to C.   Draw a clear diagram!

Two identical balls of 29.5 g are suspended from threads of length 1.1 m and carry equal charges as shown in the figure. Each ball is 1.5 cm from the centerline.a) Assume that the angle Θ is so small that its tangent can be replaced by its sine. This is called the small angle approximation and means that for small angles sin(Θ)=tan(Θ). Find the magnitude of charge on one of the balls. b) Now, assume the two balls are losing charge to the air very slowly. That means they'll be slowly approaching each other. If a ball is moving at an instantaneous speed of 2.50E−05 m/s, at what rate is the ball losing charge? Start by writing the speed of the ball and the rate of change of the charge as symbolic derivatives, and then relate those derivatives. Give your answer in Coulombs per second (C/s). Note that because the balls are losing charge so slowly, we can still use our results from the previous part, as the system is almost in equilibrium. Give your answer as a magnitude.

Two balloons have equal and opposite charges. Balloon one has N = 1010 excess electrons. The balloons are separated by d = 2.8 m and each electron has a negative charge of e = 1.602 × 10-19 C. b. What is the magnitude of the force (in N) of balloon one on balloon two using the variables provided and the Coulomb constant k (k = 8.988 × 109 N m2/C2).

Needs Complete typed solution with 100 %  accuracy.

In the figure, the point charges are located at the corners of an equilateral triangle 29 cm on a side. What is the direction of the force on qa in degrees above the negative x-axis with origin at qa?

Two charges lie in a line along the x axis. Charge 1 is q1 = 1.1 C and charge 2 is q2 = 2.15 C. They are each a distance of d = 0.093 m from the origin. What is the distance on the x-axis from the origin at which the electric field will be zero. Give your answer in meters.

Two electric charges of q1 = -4.8 nC and q2= 4 nC are separated by 3 cm. What is the Coulomb force between the two charges? How many extra electrons does q1 contain?

Energy of Interaction of Point Charges Coulomb's Law yields an expression for the energy of interaction for a pair of point charges. 2.31 x 10-1ºQ1Q2 V = V is the energy (in J) required to bring the two charges from infinite distance separation to distance r (in nm). Q1 and Q2 are the charges in terms of electrons. (i.e. the constant in the above expression is 2.31x10-19 J nm electrons-2) For a group of "point" charges (e.g. ions) the total energy of interaction is the sum of the interaction energies for the individual pairs. Calculate the energy of interaction for the square arrangement of ions shown in the diagram below. d = 0.125 nm. d (-2)