Consider two oppositely charged, isolated parallel plates separated by distance
ID: 2173283 • Letter: C
Question
Consider two oppositely charged, isolated parallel plates separated by distance D, with capacitance C, charge Q, and stored energy U. For each statement below, select "True" or "False".1.Increasing D, increases the E field.
2.Because of charge conservation, inserting a dielectric leaves C unchanged.
3. Because of energy conservation, inserting a dielectric leaves U unchanged.
4. When D is halved, Q stays the same.
5. Inserting a dielectric increases Q.
6. When D is doubled, C is halved.
7. When D is doubled, U is unchanged.
Explanation / Answer
Which statements are true for two oppositely charged, isolated parallel plates: C=capacitance, U=stored energy (Q and -Q = charge on the plates). Note: Isolated plates can not lose their charge. E = ½CV² Q = CV When the distance is doubled, C increases. false, C decreases Inserting a dielectric decreases U. inserting a dielectric increases C, which lowers V, since the charge can't change. E = ½CV². Since the V term is squared, the net energy goes down. check, assume dielectric constant doubles. C then doubles. by Q = CV, voltage halfs E = ½CV² E '= ½(2C)(V/2)² = ½(½)(C)(V)² E goes down, statement is true Inserting a dielectric increases C. true, see above When the distance is halved, Q stays the same. Q cannot change true When the distance is doubled, U increases. C = e0er(A/d) e0 is 8.8542e-12 F/m er is dielectric constant (vacuum = 1) A and d are area of plate in m² and separation in m double distance, C is cut in half. cut C in half, V doubles E = ½CV² halfing C cuts E in half, doubling V increases E by 4, so net is a doubling in energy. True Inserting a dielectric increases Q. charge doesn't change false Increasing the distance increases the Electric field capacitance goes down, charge stays constant, voltage goes up. Field is V/m, with both voltage and distance going up, so field stays the same. false