Capacitors - AP Physics 2

1. For a given $C$ and $V$, how does energy change if $V$ is doubled?

   Energy quadruples. Energy increases with the square of voltage.

2. Calculate $C_{\text{eq}}$ for $C_1 = 2\, \text{F}$, $C_2 = 3\, \text{F}$ in series.

   $C_{\text{eq}} = 1.2\, \text{F}$. Using $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$, so $C_{eq} = 1.2$ F.

3. Determine the capacitance if the charge is $3\, \text{C}$ and the voltage is $6\, \text{V}$.

   $C = 0.5\, \text{F}$. Using $C = \frac{Q}{V} = \frac{3}{6} = 0.5$ F.

4. Find $C_{\text{eq}}$ for $C_1 = 3\, \text{F}$ and $C_2 = 6\, \text{F}$ in series.

   $C_{\text{eq}} = 2\, \text{F}$. Using $\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$, so $C_{eq} = 2$ F.

5. What is the relationship between electric field $E$, voltage $V$, and plate separation $d$?

   $E = \frac{V}{d}$. Electric field is voltage divided by plate separation.

6. Determine the equivalent capacitance for two capacitors, $C_1$ and $C_2$, in parallel.

   $C_{\text{eq}} = C_1 + C_2$. Two capacitors in parallel simply add together.

7. If $Q = 5\, \text{C}$ and $V = 10\, \text{V}$, find the capacitance.

   $C = 0.5\, \text{F}$. Using $C = \frac{Q}{V} = \frac{5}{10} = 0.5 \, \text{F}$

8. State the expression for energy stored in a capacitor with charge $Q$ and voltage $V$.

   $U = \frac{QV}{2}$. Alternative form of the energy storage equation.

9. Determine the equivalent capacitance for two capacitors, $C_1$ and $C_2$, in series.

   $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$. Two capacitors in series combine using reciprocal addition.

10. State the energy stored in a capacitor formula.

    $U = \frac{1}{2}CV^2$. Energy equals half the capacitance times voltage squared.

11. What is the formula for capacitors in series?

    $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$. Series capacitors add reciprocally like resistors in parallel.

12. How does inserting a dielectric affect the electric field in a capacitor?

    Reduces the electric field. Dielectric materials weaken the field between plates.

13. Find the electric field if $V = 12\, \text{V}$ and $d = 3\, \text{m}$.

    $E = 4\, \text{V/m}$. Using $E = \frac{V}{d} = \frac{12}{3} = 4$ V/m.

14. Calculate $C_{\text{eq}}$ for $C_1 = 2\, \text{F}$, $C_2 = 3\, \text{F}$ in parallel.

    $C_{\text{eq}} = 5\, \text{F}$. Using $C_{eq} = 2 + 3 = 5$ F for parallel.

15. State the relationship between plate area and capacitance.

    Capacitance is directly proportional. Larger area provides greater capacitance linearly.

16. Explain the effect on capacitance when a dielectric material is removed.

    Capacitance decreases. Removing dielectric reduces capacitance by factor $k$.

17. How is the electric field between capacitor plates calculated?

    $E = \frac{Q}{\varepsilon_0 A}$. Field depends on charge density and permittivity.

18. Calculate the new capacitance if the dielectric constant is $k$.

    $C' = kC$. New capacitance equals dielectric constant times original.

19. What is the formula for capacitance in terms of charge and voltage?

    $C = \frac{Q}{V}$. Capacitance is the ratio of stored charge to applied voltage.

20. What is the effect of increasing plate separation on the electric field?

    Electric field decreases. Greater separation weakens the electric field strength.

21. List one factor that affects the capacitance of a parallel plate capacitor.

    Plate area or plate separation. Both area and separation directly affect capacitance.

22. What is the relationship between stored energy and voltage in a capacitor?

    $U \propto V^2$. Energy varies as the square of voltage.

23. What is the dielectric constant $k$'s role in capacitance?

    Increases capacitance by $k$ times. Dielectric constant multiplies the original capacitance value.

24. Determine the charge on a $5\, \text{F}$ capacitor with $10\, \text{V}$.

    $Q = 50\, \text{C}$. Using $Q = CV = (5)(10) = 50$ C.

25. What is the effect of a dielectric on the voltage across a capacitor?

    Voltage decreases. Dielectric reduces voltage while maintaining constant charge.

26. Describe how capacitance changes with increased plate area.

    Capacitance increases. Larger area provides more space for charge storage.

27. If $C = 4\, \text{F}$ and $V = 5\, \text{V}$, find the charge $Q$.

    $Q = 20\, \text{C}$. Using $Q = CV = (4)(5) = 20$ C.

28. What is the formula for capacitors in parallel?

    $C_{\text{eq}} = C_1 + C_2 + \cdots$. Parallel capacitors add directly like resistors in series.

29. What is the formula for energy stored in a capacitor in terms of $C$ and $Q$?

    $U = \frac{Q^2}{2C}$. Energy formula expressed in terms of charge and capacitance.

30. What happens to capacitance if plate separation increases?

    Capacitance decreases. Greater separation weakens the capacitor's ability to store charge.

31. What is the potential difference if $C = 4\, \text{F}$ and $Q = 8\, \text{C}$?

    $V = 2\, \text{V}$. Using $V = \frac{Q}{C} = \frac{8}{4} = 2$ V.

32. Identify the unit of capacitance.

    Farad (F). Named after Michael Faraday, represents capacitance measurement.

33. Determine the energy stored if $Q = 4\, \text{C}$ and $V = 5\, \text{V}$.

    $U = 10\, \text{J}$. Using $U = \frac{QV}{2} = \frac{(4)(5)}{2} = 10$ J.

34. How does the dielectric constant $k$ affect stored energy if the voltage is constant?

    Stored energy increases by $k$. Dielectric increases energy storage capacity at constant voltage.

35. What is the effect on stored energy if a dielectric is added?

    Stored energy increases. Dielectric materials enhance energy storage capability.

36. Calculate the voltage across a capacitor with $C = 2\, \text{F}$ and $Q = 10\, \text{C}$.

    $V = 5\, \text{V}$. Using $V = \frac{Q}{C} = \frac{10}{2} = 5$ V.

37. Find $C_{\text{eq}}$ for $C_1 = 3\, \text{F}$ and $C_2 = 6\, \text{F}$ in parallel.

    $C_{\text{eq}} = 9\, \text{F}$. Using $C_{eq} = 3 + 6 = 9$ F for parallel connection.

38. Find the energy stored in a capacitor with $C = 2\, \text{F}$ and $V = 3\, \text{V}$.

    $U = 9\, \text{J}$. Using $U = \frac{1}{2}CV^2 = \frac{1}{2}(2)(3^2) = 9$ J.

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