MCAT Physical : Electrostatics and Electrical Fields
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Example Questions
Example Question #1 : Other Electrical Principles
In a battery, the __________ is the site of the reduction half-reaction and the __________ is the electrode where the oxidation number of the reactant increases.
cathode . . . cathode
anode . . . anode
cathode . . . anode
anode . . . cathode
cathode . . . anode
An electrochemical cell in a battery contains two electrodes: the anode and the cathode. Anodes are the site of oxidation half-reactions and cathodes are the site of reduction half-reactions. Recall that oxidation involves loss of electrons from reactants. The reactants at the anode become more positive, and there is an increase in their oxidation number. Reduction, on the other hand, involves a gain of electrons and a decrease in the reactant oxidation number at the cathode.
Example Question #1 : Electrostatics
A neutral conducting sphere rests on an insulating styrofoam block. A student then places his hand on the sphere, maintaining this contact as a negatively charged plastic rod is brought near the sphere without touching. While the plastic rod is still nearby, the student removes his hand. Finally, the rod is also removed. Which of the following best describes the final charge on the sphere?
Net positive charge, and polarized
Zero net charge, but polarized
Net negative charge, distributed evenly over the sphere
Net positive charge, distributed evenly over the sphere
Net negative charge, and polarized
Net positive charge, distributed evenly over the sphere
The student's hand on the sphere "grounds" it, allowing charges to flow in or out. No charge flows through the styrofoam insulator. When the negatively charged rod is brought nearby, negative charges in the sphere are repelled and exit via the student's hand, giving the sphere a net positive charge. After the student removes his hand, no more charges can flow into or out from the sphere, so the net positive charge remains after the rod is removed. Since this is a conducting sphere, the charges will spread out evenly over the surface.
Example Question #2 : Electrostatics
How much energy is required by a 
Power is given in watts which is equal to
Convert 10 minutes into seconds and then multiply the time in seconds with the number of watts to obtain the joules of energy required by the lightbulb.
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