Review/Practice for Test 1 A charge q is placed at the exact center of a pyramid

Question

Review/Practice for Test 1 A charge q is placed at the exact center of a pyramidal box (a perfect tetrahedron) that is missing one of the four sides. Calculate the flux of the electric field of the charge through the remaining faces of the pyramid. Is the figure at right depicting some of the electric field lines between two conductors in electrostatic equilibrium right or wrong? Either way, explain why. What are the magnitude and direction of the electric field produced at P by the three charges in the figure at right? Determine the resistance between points A and B in the circuit at right. A fully-charged capacitor (C = 5.0 Times 104 mu F) is connected in series with a resistor (R = 1.0 M Ohm), in an open circuit. How long does it take after the switch S is closed for the charge in the capacitor to decrease to one-third its initial value? Three hollow conductors of different volumes, initially uncharged, are connected by wires as shown in the figure. A charge Q = 3.5 mu C is deposited on the big sphere. I low would the electric potentials on the objects compare to each other? Which would be the highest, if any? Seven capacitors, initially uncharged, are connected as in the figure at right. The potential difference between A and B is set to 1,000 V, and the capacitances are C1 = 1.0 pF, C2 = 2.0 pF, C3 = 3.0 pF, C4 = 4.0 pF, C5 = 5.0 pF, C6 = 6.0 pF, C7 = 7.0 pF. What is the charge on capacitor C3? A small, 2.5-g charged ball floats in equilibrium a distance d above a uniformly charged infinite sheet. If the surface charge density on the sheet is sigma = 2.1 Times 10-7 C/m, determine the charge of the ball.

Explanation / Answer

Too many question posted as one.

2) Electric field lines are not uniform on either sides, so may be the non eqauilibrium condition.

4) 2r and 3r in series, so it is 5r. Now this is in parallel with another 5r so parallel combination would be

R = (5r* 5r )/(5r+5r) = 2.5r

Now this combination is again in series with 4r so resistance here would b 2.5 r + 4 r = 6.5 r. This 6.5 r is in parallel with r so net resistance is

   Rn = (r *6.5r) /(r + 6.5r) = 0.87 r

5)  The time need for RC circuit to decrease the charge to 1/3 is

q = Q e-t / RC

1/3 = e-t / RC  

t = - RC ln (1/3)

t =- (5x10^-2)(1x10^6) ln (1/3)

t = 54930.6s ( or approximately 15.3 hrs)

6) The charge is equally distributed and all in same potential. Equipotential.

7) The potential is unchanged so Q = CV=(3x10^-12)(1000) = 3x10-9 C = 3nC

8) The force on the charge due to the electric field should be equal to the weight of the charge

F = qE = q ( sigma / epsilon) = q (2.1x10^-7) / 8.85x10^-12

F = q * 22598.9 N

now this is equal to the weight m g so

q * 22598.9 = m g

q * 22598.9 = (2.5x10^-3 )(9.8)

q = 1.08x10-6 C or 1.1uC

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.