Mastering Physics Solutions: Suspending Charged Particles Using Electric Fields

Mastering Physics Solutions: Suspending Charged Particles Using Electric Fields

On February 10, 2014, in Chapter 15: Electric Charge, Forces, and Fields, by Mastering Physics Solutions

Part A = -2.13 * 10^-5 C Click to use the calculator/solver for this part of the problem
Part B = E = 1.02 * 10^-7

What must the charge (sign and magnitude) of a particle of mass 1.43 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 660 N/C?
What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

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Mastering Physics Solutions: An Electron in a Diode

Mastering Physics Solutions: An Electron in a Diode

On February 9, 2014, in Chapter 16: Electric Potential, Energy, and Capacitance, by Mastering Physics Solutions

Part A = 9.16 * 10^6 m/s

Find its speed vfinal when it strikes the anode.

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Mastering Physics Solutions: Batteries in Series or Parallel

Mastering Physics Solutions: Batteries in Series or Parallel

On November 29, 2013, in Chapter 18: Basic Electric Circuits, by Mastering Physics Solutions

Part A = counterclockwise
Part B = I = 2EMF / ((2 * R1) + R2)
Part C = IB = 2*EMF / (R1 + 2*R2)
Part D = 0.064 W
Part E = R1 / R2 = 1
Part F = R2 > R1

In which direction does the current in circuit A flow?
What is the current through the resistor of resistance R2 in circuit A?
Calculate the current IB through the resistor of resistance R2 for circuit B.
What is the power dissipated by the resistor of resistance R2 for circuit A, given that Ε = 10 V, R1 = 300 ohms, and R2 = 5000 ohms?
For what ratio of R1 and R2 would power dissipated by the resistor of resistance R2 be the same for circuit A and circuit B?
Under which of the following conditions would power dissipated by the resistance R2 in circuit A be bigger than that of circuit B?

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Mastering Physics Solutions: Kirchhoff’s Rules and Applying Them

Mastering Physics Solutions: Kirchhoff’s Rules and Applying Them

On September 23, 2013, in Chapter 18: Basic Electric Circuits, by Mastering Physics Solutions

Part A = current
Part B = I2 + I3 – I1
Part C = I3 ⋅ R3 – I2 ⋅ R2
Part D = Vb – I1 ⋅ R1 – I3 ⋅ R3

The junction rule describes the conservation of which quantity? Note that this rule applies only to circuits that are in a steady state.
Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance R2).
Apply the loop rule to loop 2 (the smaller loop on the right). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow. Remember that the current meter is ideal.
Now apply the loop rule to loop 1 (the larger loop spanning the entire circuit). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow.

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Mastering Physics Solutions: Electrostatic Force of Water on a Chlorine Ion

Mastering Physics Solutions: Electrostatic Force of Water on a Chlorine Ion

On September 15, 2013, in Chapter 15: Electric Charge, Forces, and Fields, by Mastering Physics Solutions

Part A = 6.58 * 10^-13 N
Part B = negative x
Part C = attractive

Find the magnitude of the electric force, ignoring the sign, that the water molecule exerts on the chlorine ion.
What is the direction of the electric force?
Is this force attractive or repulsive?

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