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I'm having a hard time understanding the concept and I'd like to chat with you in the comments about your methods. Please write big and legibly when showing your work.

I'm having a hard time understanding the concept and I'd like to chat with you in the comments about your methods. Please write big and legibly when showing your work. P 3. Two equal-magnitude point charges, and Q. are fixed in space as shown above, forming an equal dipole. Five equipotential sufaces (dashed curves) are drawn for you in the diagram and labeled with their voltages. (Assume that V-0 at r- away from the charges.) a. Draw between 8 and 12 electric field lines on the diagram above. Include directional arrewbeads on ALL of your field lines. b. Suppose that an electron is released from rest at point P. Immediately afterward, the electron experiences a net electric foree in which direction? C. toward P D, towardP E zero net force A. toward P B, soward P c. Suppose that a quadrup ly ionized silicon ion(Si·') is released from rest at point Pi Immediately afterward the ion experiences a net electric force in which direction? A. toward Ps B. away fromP C. toward OA D. away from E to the left d. Suppose that a proton is moved from a point located on the-500, volt equipotential line to a point on the 500 volt equipotential line. (The proton starts and ends at rest.) What is the proton's change in potential energy? D. 500.e E. +1000, eV e. Suppose you move a quadruply ionized silicon ion (Si) from point P to point P.. (The ion starts and ends at rest.) How much work must you perfoem on the ion? A. 250, e . 500 e C. 1000 e D. 2000, e E 4000. ev L Moving a proton from point Pz to point P, roquires point P, to point P·(The proton starts and ends at rest in either case.) energy than moving a proton from B. the same C. less D. cannot determine from information givern

Explanation / Answer

answering 4 parts ( chegg's policy)

b) towards P4 ( owing to force of attraction)

c)towards Qa ( owing to force of attraction)

d)chnage in potential energy = q ( Va-Vb) = 1.6 x 10^-19 (500 +500) J = 1.6 x 10^-19 x 10^ 3 J = 9.9872 x 10^ 2 eV = 1000 eV apprx

e)work don e= change in PE = 4 x 1.6 x 10^-19 (500-0) = 32 x 10^-17 J = 199.744 x 10 eV = 1997.44 eV = 2000 eV apprx]

f) same energy

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