1 of 4CTVoltage-0.Two charges, +Q and +2Q are released from rest a distance R apart. Each particle feels only the coulomb force due to the other charge. As the particles continue to move apart after their release, the speed of each particleA) decreases B) increases C) stays the sameAnswer: the speed increases. The acceleration decreases as they move apart, but the acceleration is never zero, so the particles never stop going faster and faster.CTVoltage-1 The work done by a constant force Fr in moving an object through a displacementrDr is defined as FW F r= �DrrA person carefully lowers a book at constant speed a distance h. The work done by the person, done by gravity, and done by the netforce on the book are:person gravity net forceA) + – +B) – + –C) – + 0D) + – 0E) None of these.Answer: The work done by the person is negative, the work done by gravity is positive, the work done by the net force is zero.1/14/2019 PHYS1120 Dubson Fa09 ©University of Coloradoparticle 1+QRparticle 2+2QFrinitial final(f)h(i)2 of 4CTVoltage-2 A positive test charge +q is carefullymoved by some external agent (tweezers) at constantspeed a distance x between two capacitor plates in thedirection along the electric field. The work done by the agent, done by the electric field,and done by the net force on the test charge are:agent field net forceA) + – +B) – + –C) – + 0D) + – 0E) None of these.The change in electrostatic potential energy PEext fieldU W WD =+ =- of positive test charge wasA) +(PE increased) B) – (PE decreased) C) 0 (PE constant)The change in voltage of the test charge wasA) negative(V decreased) B) positive (V increased) C) zero, no change in voltageAnswers: The work done by the agent is negative (the agent has to restrain the particle as it is moved). The work done by the field is positive, the work done by the net force is zero.The change in the PE was negative (PE became more negative), since the work done by the external agent is negative.The change in the voltage is negative (V became more negative as we moved from near +charges to near – charges.)1/14/2019 PHYS1120 Dubson Fa09 ©University of ColoradoE+qx3 of 4CTVoltage-3 A positive charge +q is moved from position i to position f between the plates of a charged capacitor as shown. As the test charge +q was moved from i to f, the potential energy (U) increased or decreased and the voltage (V) at the position of the test charge increased or decreased.A) PE increased and V decreased.B) PE decreased and V increased.C) PE increased and V increased.D) PE decreased and V decreased.E) None of these.What if the test charge was negative –q (same choices) ?Answers: For the positive charge (+q) moving to the left, the PE increases and the voltage V also increases. For a negative charge (–q) moving to the left, the PE decreases,but the voltage still increases. The voltage at the location of a test charge does not depend on the test charge. The PE ofa test charge does depend on the test charge.1/14/2019 PHYS1120 Dubson Fa09 ©University of ColoradoxE+q(i)(f)4 of 4CTVoltage-4Consider 4 charges +Q, +Q, –Q, and –Q arranged in a square, with points X and Y located midway between a pair of charges, as shown. At point X, the voltage is.. A) positive B) negative C) zeroAt point Y, the voltage is.. A) positive B) negative C) zeroAt point Y, the electric field ...A) is zero B) points right C) points leftD) points up E) points downAnswers: At point X, the voltage is zero. At point Y, the voltage is positive. At Point Y, the E-field points right.1/14/2019 PHYS1120 Dubson Fa09 ©University of Colorado+Q+Q
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