Capacitors - AP Physics C Electricity and Magnetism

1. A parallel-plate capacitor creates a uniform electric field of magnitude $E$ in the volume between its plates. The permittivity of free space is $\epsilon_0$. What is the energy density, or energy stored per unit volume, in the electric field?
   Answer: $\frac{1}{2}\epsilon_0 E^2$ The energy stored in a capacitor is $U = \frac{1}{2}CV^2$. For a parallel-plate capacitor, $C = \epsilon_0 A/d$ and $V = Ed$. Substituting these gives $U = \frac{1}{2}(\frac{\epsilon_0 A}{d})(Ed)^2 = \frac{1}{2}\epsilon_0 A d E^2$. The volume between the plates is $Vol = Ad$. The energy density $u_E$ is $U/Vol = \frac{\frac{1}{2}\epsilon_0 A d E^2}{Ad} = \frac{1}{2}\epsilon_0 E^2$.
2. An electron with initial velocity $v_0$ enters a region of uniform electric field $E$ directed perpendicular to its velocity. The region is between the plates of a large parallel-plate capacitor. Neglecting any edge effects and gravitational forces, which of the following best describes the path of the electron while it is in the electric field?
   Answer: A parabolic arc. The electron experiences a constant force $F = -eE$ in the direction opposite to the electric field. This force is perpendicular to its initial velocity. According to Newton's second law, this results in a constant acceleration $a = F/m_e$ in that direction. The situation is analogous to projectile motion under gravity, where there is constant velocity in one direction and constant acceleration in the perpendicular direction. The resulting path is a parabola.
3. The electric potential energy $U$ of a uniformly charged conducting sphere of radius $R$ and total charge $Q$ is $U = \frac{Q^2}{8\pi\epsilon_0 R}$. This can be interpreted as the energy stored in a capacitor. What is the effective self-capacitance of the sphere?
   Answer: $4\pi\epsilon_0 R$ The energy stored in a capacitor is given by the formula $U = \frac{Q^2}{2C}$. We are given the energy of the charged sphere as $U = \frac{Q^2}{8\pi\epsilon_0 R}$. By comparing these two expressions, we can find the capacitance $C$. Setting them equal: $\frac{Q^2}{2C} = \frac{Q^2}{8\pi\epsilon_0 R}$. Solving for $C$ gives $2C = 8\pi\epsilon_0 R$, so $C = 4\pi\epsilon_0 R$. This corresponds to the capacitance of an isolated sphere.
4. An uncharged capacitor is connected to a battery. The process of charging involves moving a total charge $Q_f$ from one plate to the other, resulting in a final potential difference $V_f$. The total work required to charge the capacitor is $W$. What is the average potential difference through which each infinitesimal amount of charge is moved during the charging process?
   Answer: $V_f/2$ The work done to charge a capacitor is equal to the energy stored, $W = U = \frac{1}{2}Q_f V_f$. The work can also be thought of as the total charge moved multiplied by the average potential difference, $W = Q_f V_{avg}$. Equating the two expressions for work gives $Q_f V_{avg} = \frac{1}{2}Q_f V_f$. Solving for the average potential difference yields $V_{avg} = V_f/2$. This reflects that the potential difference increases linearly from 0 to $V_f$ as the capacitor is charged.
5. A parallel-plate capacitor with capacitance $C_0$ is charged to potential difference $V_0$ and then disconnected from the battery. A dielectric slab with dielectric constant $\kappa > 1$ is then inserted, completely filling the space between the plates. How does the final stored energy $U_f$ compare to the initial stored energy $U_i$?
   Answer: $U_f = U_i / \kappa$ Since the capacitor is disconnected from the battery, the charge $Q_0$ on its plates remains constant. The initial energy is $U_i = Q_0^2 / (2C_0)$. When the dielectric is inserted, the capacitance becomes $C_f = \kappa C_0$. The final energy is $U_f = Q_0^2 / (2C_f) = Q_0^2 / (2\kappa C_0) = (1/\kappa) (Q_0^2 / (2C_0)) = U_i / \kappa$. Since $\kappa > 1$, the stored energy decreases.
6. A parallel-plate capacitor is held with its plates horizontal. It is charged and then disconnected from the battery. A slab of dielectric material is placed on the bottom plate and released. The slab is pulled into the region between the plates. Which of the following is a correct explanation for this phenomenon?
   Answer: The dielectric is pulled in to a position of lower potential energy, which occurs because inserting it increases the capacitance. The system moves to a state of lower potential energy. For an isolated capacitor, the energy is $U = Q^2/(2C)$. When the dielectric is inserted, the capacitance $C$ increases. Since $Q$ is constant, the stored energy $U$ decreases. The decrease in potential energy is converted into the kinetic energy of the slab and the work done by the electric field, indicating an attractive force that pulls the slab in.
7. A long cylindrical capacitor has an inner conductor of radius $a$ and an outer conductor of radius $b$. The inner conductor holds a charge per unit length of $+\lambda$. The electric field in the region $a < r < b$ is given by $E = \frac{\lambda}{2\pi\epsilon_0 r}$. What is the magnitude of the potential difference between the conductors?
   Answer: $\frac{\lambda}{2\pi\epsilon_0} \ln(b/a)$ The potential difference $V_{ab}$ is found by integrating the electric field from the inner to the outer conductor: $V_{ab} = -\int_a^b \vec{E} \cdot d\vec{l}$. Since $E$ is directed radially outward, we integrate along a radial path. $V_{ab} = V_a - V_b = \int_a^b E dr = \int_a^b \frac{\lambda}{2\pi\epsilon_0 r} dr = \frac{\lambda}{2\pi\epsilon_0} [\ln(r)]_a^b = \frac{\lambda}{2\pi\epsilon_0} (\ln(b) - \ln(a)) = \frac{\lambda}{2\pi\epsilon_0} \ln(b/a)$.
8. A parallel-plate capacitor is charged by a battery to store energy $U_0$. The capacitor is then disconnected from the battery. The distance between the plates is then doubled. What is the new energy $U_f$ stored in the capacitor?
   Answer: $2U_0$ When the capacitor is disconnected, the charge $Q$ on the plates is conserved. The capacitance $C$ is inversely proportional to the plate separation $d$. Doubling $d$ halves the capacitance, so $C_f = C_0/2$. The energy stored in a capacitor can be expressed as $U = Q^2/(2C)$. Since $Q$ is constant and $C$ is halved, the new energy $U_f = Q^2/(2(C_0/2)) = Q^2/C_0 = 2(Q^2/(2C_0)) = 2U_0$.
9. A parallel-plate capacitor with charge $Q_0$ is fully discharged by connecting its terminals with a wire of resistance $R$. The total energy $U_0$ initially stored in the capacitor is dissipated as heat in the wire. If the same capacitor were instead charged to $3Q_0$, what would be the total energy dissipated in the wire during its discharge?
   Answer: $9U_0$ The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$. This is the energy that will be dissipated during discharge. The initial energy with charge $Q_0$ is $U_0 = \frac{Q_0^2}{2C}$. If the capacitor is charged to $3Q_0$, the new stored energy is $U_{new} = \frac{(3Q_0)^2}{2C} = \frac{9Q_0^2}{2C} = 9U_0$. By conservation of energy, this is the total energy that will be dissipated.
10. The energy stored in the electric field of a charged system can be calculated by integrating the energy density $u_E = \frac{1}{2} \epsilon_0 E^2$ over all space. For an isolated conducting sphere of radius $R$ and charge $Q$, the electric field for $r > R$ is $E = \frac{Q}{4\pi\epsilon_0 r^2}$. Which integral correctly calculates the energy stored in the field outside the sphere?
    Answer: $\int_R^{\infty} \frac{1}{2} \epsilon_0 \left(\frac{Q}{4\pi\epsilon_0 r^2}\right)^2 (4\pi r^2) dr$ To find the total energy, we must integrate the energy density $u_E$ over the volume of space where the field exists. The volume element $dV$ for a spherical shell of radius $r$ and thickness $dr$ is $dV = 4\pi r^2 dr$. The integration must be performed over the region outside the sphere, so the limits are from $r=R$ to $r=\infty$. The integrand is $u_E dV$. Therefore, the correct integral is $U = \int_{V} u_E dV = \int_R^{\infty} \frac{1}{2} \epsilon_0 E^2 (4\pi r^2) dr = \int_R^{\infty} \frac{1}{2} \epsilon_0 \left(\frac{Q}{4\pi\epsilon_0 r^2}\right)^2 (4\pi r^2) dr$.
11. An isolated parallel-plate capacitor has plates of area $A$ separated by a distance $d$. It holds a net charge of magnitude $Q$ on each plate. What is the magnitude of the electrostatic force exerted by one plate on the other?
    Answer: $\frac{Q^2}{2\epsilon_0 A}$ The electric field produced by one plate alone is $E_1 = \frac{\sigma}{2\epsilon_0} = \frac{Q}{2\epsilon_0 A}$. The force exerted by this field on the other plate, which has charge $Q$, is $F = Q E_1 = Q \left(\frac{Q}{2\epsilon_0 A}\right) = \frac{Q^2}{2\epsilon_0 A}$. Using the total field between the plates ($E = \sigma/\epsilon_0$) is incorrect because a plate does not exert a force on itself.
12. A coaxial cylindrical capacitor has a length $L$, an inner conductor of radius $a$, and an outer conductor of radius $b$. Its capacitance is given by $C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}$. If the inner radius $a$ is halved and the outer radius $b$ is doubled, how does the new capacitance $C'$ compare to the original capacitance $C$?
    Answer: $C' = \frac{\ln(b/a)}{\ln(4b/a)} C$ The original capacitance is $C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}$. The new radii are $a' = a/2$ and $b' = 2b$. The new capacitance is $C' = \frac{2\pi\epsilon_0 L}{\ln(b'/a')} = \frac{2\pi\epsilon_0 L}{\ln((2b)/(a/2))} = \frac{2\pi\epsilon_0 L}{\ln(4b/a)}$. The ratio $C'/C = \frac{\ln(b/a)}{\ln(4b/a)}$. Therefore, $C' = \frac{\ln(b/a)}{\ln(4b/a)} C$.
13. A parallel-plate capacitor with capacitance $C$ is connected to a battery with a constant potential difference $V$. While the capacitor is connected to the battery, a dielectric slab with dielectric constant $\kappa$ is inserted between the plates. What is the magnitude of the work done by the battery during this process?
    Answer: $(\kappa - 1) C V^2$ The work done by the battery is $W_{batt} = (\Delta Q)V$, where $\Delta Q$ is the additional charge that flows from the battery. Initially, the charge is $Q_i = CV$. After inserting the dielectric, the capacitance becomes $C_f = \kappa C$, and the final charge is $Q_f = C_f V = \kappa C V$. The additional charge that flows is $\Delta Q = Q_f - Q_i = (\kappa - 1)CV$. Therefore, the work done by the battery is $W_{batt} = (\kappa - 1)CV \cdot V = (\kappa - 1)CV^2$.
14. A parallel-plate capacitor is connected to a battery with a constant potential difference $V_0$. While connected, the area of the plates that directly overlaps is decreased by a factor of two. What is the ratio of the final stored energy to the initial stored energy?
    Answer: $1/2$ While connected to the battery, the potential difference $V_0$ remains constant. The initial energy is $U_i = \frac{1}{2}C_i V_0^2$. The capacitance $C$ is proportional to the plate area $A$. Halving the area halves the capacitance, so $C_f = C_i/2$. The final energy is $U_f = \frac{1}{2}C_f V_0^2 = \frac{1}{2}(\frac{C_i}{2})V_0^2 = \frac{1}{2} U_i$. The ratio $U_f/U_i$ is $1/2$.
15. The electric field between the plates of a charged parallel-plate capacitor is nearly uniform, but there are non-uniform 'fringing fields' near the edges. The formula $C = \epsilon_0 A/d$ is derived assuming a uniform field. How does the actual capacitance of a real capacitor compare to the value predicted by this formula?
    Answer: The actual capacitance is greater, because the fringing fields store additional energy. Capacitance is defined as $C=Q/V$. For a given charge $Q$, the fringing fields mean that the electric field lines occupy a larger effective volume, extending beyond the edges of the plates. This effectively stores more energy for the same amount of charge compared to the ideal case. A greater storage capacity for a given charge implies a greater capacitance. Equivalently, for a given voltage V, more charge can be stored due to the fringing fields, thus increasing capacitance.
16. A spherical capacitor consists of two concentric conducting spheres of radii $a$ and $b$, with $b>a$. Its capacitance is given by $C = 4\pi\epsilon_0 \frac{ab}{b-a}$. What expression is obtained for the capacitance in the limit that the outer sphere's radius $b$ approaches infinity, representing an isolated sphere of radius $a$?
    Answer: $4\pi\epsilon_0 a$ To find the limit as $b \to \infty$, we can rewrite the expression as $C = 4\pi\epsilon_0 \frac{a}{1 - a/b}$. As $b \to \infty$, the term $a/b$ approaches 0. Therefore, the expression for the capacitance approaches $C = 4\pi\epsilon_0 \frac{a}{1 - 0} = 4\pi\epsilon_0 a$. This is the capacitance of an isolated conducting sphere of radius $a$.
17. A capacitor with capacitance $C$ is charged to a potential difference $V$, storing an amount of energy $U$. If the potential difference is increased to $3V$, what is the new amount of energy stored in the capacitor?
    Answer: $9U$ The energy $U$ stored in a capacitor is given by $U = \frac{1}{2}C V^2$. Since the energy is proportional to the square of the potential difference, tripling the voltage from $V$ to $3V$ will increase the stored energy by a factor of $3^2 = 9$. The new energy will be $9U$.
18. An air-filled parallel-plate capacitor has a capacitance of $C_0$. If the area of the plates is tripled and the distance between the plates is halved, what is the new capacitance?
    Answer: $6 C_0$ The capacitance of a parallel-plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$, where $A$ is the area of the plates and $d$ is the separation distance. The new area is $A' = 3A$ and the new separation is $d' = d/2$. The new capacitance $C'$ is $C' = \frac{\epsilon_0 A'}{d'} = \frac{\epsilon_0 (3A)}{d/2} = 6 \frac{\epsilon_0 A}{d} = 6C_0$.

Start this quiz