Capacitors
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AP Physics C: Electricity and Magnetism › Capacitors
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?
$$6 C_0$$
$$3 C_0$$
$$\frac{2}{3} C_0$$
$$\frac{3}{2} C_0$$
Explanation
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 $$C' = \frac{\epsilon_0 A'}{d'} = \frac{\epsilon_0 (3A)}{d/2} = 6 \frac{\epsilon_0 A}{d} = 6C_0$$.
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?
A circular arc.
A parabolic arc.
A straight line in the direction of $$E$$.
A straight line in the direction of $$v_0$$.
Explanation
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.
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?
$$U_0$$
$$9U_0$$
$$3U_0$$
$$U_0/3$$
Explanation
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.
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?
$$\frac{1}{2} (\kappa - 1) C V^2$$
$$\frac{1}{2} (1 - 1/\kappa) C V^2$$
$$(\kappa - 1) C V^2$$
$$(1 - 1/\kappa) C V^2$$
Explanation
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 $$W_{batt} = (\kappa - 1)CV \cdot V = (\kappa - 1)CV^2$$.
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?
$$\frac{\lambda}{2\pi\epsilon_0} (1/a - 1/b)$$
$$\frac{\lambda}{2\pi\epsilon_0} \ln(b/a)$$
$$\frac{\lambda b}{2\pi\epsilon_0 a}$$
$$\frac{\lambda}{2\pi\epsilon_0} (b-a)$$
Explanation
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. $$V_{ab} = V_a - V_b = \int_a^b E dr = \int_a^b \frac{\lambda}{2\pi\epsilon_0 r} dr = \frac{\lambda}{2\pi\epsilon_0} [\ln(r)]_a^b = \frac{\lambda}{2\pi\epsilon_0} (\ln(b) - \ln(a)) = \frac{\lambda}{2\pi\epsilon_0} \ln(b/a)$$.
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?
$$\epsilon_0 E^2$$
$$\frac{1}{2}\epsilon_0 E^2$$
$$\frac{E^2}{2\epsilon_0}$$
$$\frac{\epsilon_0 E}{2}$$
Explanation
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 $$U/Vol = \frac{\frac{1}{2}\epsilon_0 A d E^2}{Ad} = \frac{1}{2}\epsilon_0 E^2$$.
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?
$$1/4$$
$$1/2$$
$$4$$
$$2$$
Explanation
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 $$1/2$$.
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?
$$\int_R^{\infty} \frac{1}{2} \epsilon_0 \left(\frac{Q}{4\pi\epsilon_0 r^2}\right)^2 (4\pi r^2) dr$$
$$\int_R^{\infty} \frac{1}{2} \epsilon_0 \left(\frac{Q}{4\pi\epsilon_0 r^2}\right) (4\pi r^2) dr$$
$$\int_R^{\infty} \frac{1}{2} \epsilon_0 \left(\frac{Q}{4\pi\epsilon_0 r^2}\right)^2 dr$$
$$\int_0^R \frac{1}{2} \epsilon_0 \left(\frac{Q}{4\pi\epsilon_0 r^2}\right)^2 (4\pi r^2) dr$$
Explanation
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 $$U = \int_{V} u_E dV = \int_R^{\infty} \frac{1}{2} \epsilon_0 E^2 (4\pi r^2) dr = \int_R^{\infty} \frac{1}{2} \epsilon_0 \left(\frac{Q}{4\pi\epsilon_0 r^2}\right)^2 (4\pi r^2) dr$$.
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?
$$3U$$
$$9U$$
$$U/3$$
$$6U$$
Explanation
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 $$9U$$.
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?
$$\frac{Q^2 d}{\epsilon_0 A}$$
$$\frac{Q^2}{2\epsilon_0 A}$$
$$\frac{Q^2}{2\epsilon_0 A d}$$
$$\frac{Q^2}{\epsilon_0 A}$$
Explanation
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.