Show all work to receive all credit. Place your work in a blue book. 1. Two char
ID: 1790931 • Letter: S
Question
Show all work to receive all credit. Place your work in a blue book. 1. Two charges are placed at a distance of 20.5 cm from each other. One charge has a magnitude three times that of the other. If the force of each charge on the other is 30N then what is the magnitude of the charges? 2. A point charge of-3.00 uC is fixed at the origin. A second +2.00 uC charge is fixed on the x-axis at 0.100m. At what points along the x-axis is the net electric field zero? 3. A cubical surface encloses a point charge of 30.00 × 10-12 C. What is the electrical flux through each surface? 4. A 12 mm long glass rod has a charge per unit length of +0.00500 C/m. How many electrons have been removed from the rod in order to produce this charge? 5. Suppose that 9.5 x 10-20 J of work is needed to take one sodium ion (NA") from the interior of a cell across its me mbrane to the outside of the cell. What is the potential difference between the inner and outer surfaces of the cell? (Use qo = 1.609 × 10-19C) 6. A 30.0 V battery of negligible internal resistance is attached to resistors Ri = 4.00 and R2 = Draw the appropriate circuit diagram and determine the power delivered by the battery and the power absorbed by each resistor if (a) they are in series and (b) if they are in parallel. 7. A 50.0 V battery of negligible internal resistance is attached to resisters R1 = 2.00 and R,-4.00 . Draw the appropriate circuit diagram and determine the power delivered by the battery and the power absorbed by each resistor if (a) they are in series and (b) if they are in parallel. 8. Find (a) the equivalent resistance of the entire circuit (b) the total current flowing through the entire circuit and (c) the current over each resistor for the circuit shown below and (d) the voltage drop across each resistor. Each resistor is 1 60 240 VExplanation / Answer
1)Let one charge be q then the other one will be 3q
F=kq1q2/r^2
F=3kq^2/r^2
q^2=F*r^2/3k
q^2=30*(.205)^2/3*9*10^{9}=46.7pC
q=6.83 *10^{-6} C
and the other is
q2=3q=3*(6.83 *10^{-6})=20.49*10^{-6} C