Solution: Electric potential at a point is defined as the amount of work done in bringing a unit positive test charge from infinity to that point against the electrostatic force. It is a scalar quantity.
Solution: Electrostatic Potential (or Potential Difference).
Solution: Given $q = 4 \times 10^{-6} \text{ C}$, $V = 10^4 \text{ V}$. Work done, $W = qV = (4 \times 10^{-6}) \times 10^4 = 0.04 \text{ Joules}$.
Solution: Potential difference, $\Delta V = \frac{W}{q} = \frac{6 \text{ J}}{3 \text{ C}} = 2 \text{ Volts}$.
Solution: Inside a conductor, electric field $E = 0$. Since $E = -dV/dr = 0$, it implies $dV = 0$. Hence, potential $V$ is constant throughout the volume and equals the value on the surface.
Solution: Yes. Example: At any point on the equatorial line of an electric dipole, the potential is zero but the electric field is not zero (it is directed anti-parallel to the dipole moment).
Solution: Yes. Example: Inside a charged spherical hollow conductor, the electric field is zero everywhere, but the electric potential is constant and non-zero (equal to surface potential).
Solution: $E = 20 \text{ V/m}$ (downward). Moving up means moving opposite to E-field. $dV = -E \cdot dr$. Here, $\Delta V = E \times d = 20 \times 0.50 \text{ m} = 10 \text{ V}$.
Solution: $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r} = (9 \times 10^9) \times \frac{4 \times 10^{-7}}{0.09} = \frac{36 \times 10^2}{0.09} = 4 \times 10^4 \text{ V}$.
Solution: Work $W = qV = (2 \times 10^{-9}) \times (4 \times 10^4) = 8 \times 10^{-5} \text{ J}$. No, work done is independent of the path because electrostatic force is a conservative force.
Solution: The total electrostatic potential at a point due to a system of point charges is the algebraic sum of the potentials due to individual charges at that point. $V_{net} = V_1 + V_2 + \dots + V_n$.
Solution: Let test charge $q_0$ be brought from $\infty$ to $P$ at distance $r$. Force $F = \frac{1}{4\pi\epsilon_0}\frac{q q_0}{x^2}$. Work done for small displacement $dx$ is $dW = F \cdot (-dx)$. Total work $W = -\int_{\infty}^{r} \frac{1}{4\pi\epsilon_0}\frac{q q_0}{x^2} dx = \frac{q q_0}{4\pi\epsilon_0} \left[ \frac{1}{x} \right]_{\infty}^{r} = \frac{1}{4\pi\epsilon_0}\frac{q q_0}{r}$. Hence $V = W/q_0 = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$.
Solution: Since $V \propto 1/r$, the graph is a rectangular hyperbola in the first quadrant, curving downwards as $r$ increases.
Solution: Let point $P$ be at distance $x$ cm from $3 \times 10^{-8} \text{ C}$ charge. Then distance from $-2 \times 10^{-8} \text{ C}$ is $(15-x)$. Net $V = \frac{k(3 \times 10^{-8})}{x} + \frac{k(-2 \times 10^{-8})}{15-x} = 0$. $\Rightarrow \frac{3}{x} = \frac{2}{15-x} \Rightarrow 45 - 3x = 2x \Rightarrow 5x = 45 \Rightarrow x = 9 \text{ cm}$.
Solution: Diagonal of square = $\sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{4} = 2 \text{ m}$. Distance from corner to center $r = 1 \text{ m}$. $V_{net} = \frac{k}{r} (q_1 + q_2 + q_3 + q_4) = \frac{9 \times 10^9}{1} (2 + 1 - 2 + 3) \times 10^{-9} = 9 \times 4 = 36 \text{ V}$.
Solution: Zero. Any point on the equatorial line is equidistant from both $+q$ and $-q$. $V = V_+ + V_- = \frac{kq}{r} + \frac{k(-q)}{r} = 0$.
Solution: $V = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$ (where $p$ is the dipole moment).
Solution: $V = \frac{kp}{r^2} = \frac{9 \times 10^9 \times 4 \times 10^{-9}}{(0.3)^2} = \frac{36}{0.09} = 400 \text{ V}$.
Solution: $V = \frac{1}{4\pi\epsilon_0} \frac{q}{R} = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{0.10} = \frac{45 \times 10^3}{0.1} = 4.5 \times 10^5 \text{ V}$.
Solution: Distance from center to each vertex $r = 10 \text{ cm} = 0.1 \text{ m}$. $V_{total} = 6 \times \frac{kq}{r} = 6 \times \frac{9 \times 10^9 \times 5 \times 10^{-6}}{0.1} = 6 \times 4.5 \times 10^5 = 2.7 \times 10^6 \text{ V}$.
Solution: An equipotential surface is a surface over which the electric potential is constant at every point.
Solution: They are concentric spherical shells centered at the location of the positive point charge.
Solution: They are a series of parallel planes perpendicular to the z-axis (i.e., planes parallel to the x-y plane).
Solution: If they intersected, the point of intersection would have two different values of electric potential simultaneously, which is impossible.
Solution: Work done to move a charge $q$ on an equipotential surface is zero ($W = q\Delta V = 0$). Also, $W = \vec{F} \cdot d\vec{r} = q\vec{E} \cdot d\vec{r} = q E dr \cos\theta$. Since $W=0$ and $E, dr \neq 0$, $\cos\theta = 0$, so $\theta = 90^\circ$. Hence $\vec{E} \perp d\vec{r}$.
Solution: [Student must draw: Spheres around each charge that get squashed closer to each other in the middle, and a straight vertical plane exactly halfway between them where $V=0$].
Solution: Work done $dW = \vec{F} \cdot d\vec{r} = (q_0\vec{E}) \cdot d\vec{r} = q_0 E dr \cos(180^\circ) = -q_0 E dr$. By definition, $dW = q_0 dV$. Equating them: $q_0 dV = -q_0 E dr \Rightarrow E = -dV/dr$.
Solution: $\vec{E} = -300\hat{i}$. $V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{r} = -\int_{0.5}^{1.2} (-300) dx = 300 [x]_{0.5}^{1.2} = 300(1.2 - 0.5) = 300(0.7) = 210 \text{ V}$.
Solution: The surface of a conductor is an equipotential surface. Therefore, the tangential component of the electric field must be zero (otherwise work would be done moving charges on the surface). Thus, $\vec{E}$ must be strictly normal.
Solution: $E = -\frac{dV}{dx} = -\frac{d}{dx}(5x^2 + 10x - 9) = -(10x + 10)$. At $x = 1$, $E = -(10(1) + 10) = -20 \text{ V/m}$. Magnitude is $20 \text{ V/m}$.
Solution: It is defined as the total amount of work done in assembling the system of charges by bringing them from infinity to their present locations.
Solution: Increases. Positive charges repel each other, so external work must be done against the electrostatic force to bring them closer, which gets stored as increased potential energy.
Solution: $U = U_{12} + U_{23} + U_{31} = \frac{kq^2}{a} + \frac{kq^2}{a} + \frac{kq^2}{a} = \frac{3kq^2}{a} = \frac{3}{4\pi\epsilon_0}\frac{q^2}{a}$.
Solution: $U = \frac{k q_1 q_2}{r} = \frac{(9 \times 10^9) \times (4 \times 10^{-6}) \times (-2 \times 10^{-6})}{1} = -72 \times 10^{-3} \text{ J} = -0.072 \text{ J}$.
Solution: $U = \frac{k (e)(-e)}{r} = -\frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{0.53 \times 10^{-10}} \text{ Joules}$. In $\text{eV}$, divide by $1.6 \times 10^{-19}$. $U = -\frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{0.53 \times 10^{-10}} \text{ eV} = -27.2 \text{ eV}$.
Solution: $U = -pE \cos\theta = -\vec{p} \cdot \vec{E}$, where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$.
Solution: (i) Minimum when $\theta = 0^\circ$ ($U = -pE$, stable equilibrium). (ii) Maximum when $\theta = 180^\circ$ ($U = +pE$, unstable equilibrium).
Solution: Torque $\tau = pE\sin\theta$. Work $W = \int_{\theta_1}^{\theta_2} \tau d\theta = \int_{\theta_1}^{\theta_2} pE\sin\theta d\theta = pE [-\cos\theta]_{\theta_1}^{\theta_2} = pE(\cos\theta_1 - \cos\theta_2)$.
Solution: $\theta_1 = 0^\circ$, $\theta_2 = 180^\circ$. $W = pE(\cos 0^\circ - \cos 180^\circ) = pE(1 - (-1)) = 2pE$. $W = 2 \times (3 \times 10^{-8}) \times 10^4 = 6 \times 10^{-4} \text{ J}$.
Solution: $U = \frac{k q_1 q_2}{r} \Rightarrow -0.5 \times 10^{-6} = \frac{9 \times 10^9 \times 5 \times 10^{-9} \times (-2 \times 10^{-9})}{r}$. $-0.5 \times 10^{-6} = -\frac{90 \times 10^{-9}}{r} \Rightarrow r = \frac{90 \times 10^{-9}}{0.5 \times 10^{-6}} = 180 \times 10^{-3} = 0.18 \text{ m} = 18 \text{ cm}$. Distance is $|x - 2| = 18$. So $x - 2 = 18 \Rightarrow x = 20 \text{ cm}$, or $2 - x = 18 \Rightarrow x = -16 \text{ cm}$.
Solution: Capacitance is the ability of a conductor to store electric charge. $C = Q/V$. SI unit is Farad (F).
Solution: $V = \frac{1}{4\pi\epsilon_0}\frac{q}{R} \Rightarrow R = \frac{q}{4\pi\epsilon_0 V} = \frac{9 \times 10^9 \times 10^{-12}}{100} = 9 \times 10^{-5} \text{ m} = 0.09 \text{ mm}$.
Solution: (i) Area of the plates ($A$), (ii) Distance of separation between plates ($d$), and (iii) Nature of the dielectric medium between them.
Solution: Electric field between plates $E = \sigma/\epsilon_0 = Q/(A\epsilon_0)$. Potential difference $V = Ed = Qd/(A\epsilon_0)$. Capacitance $C = Q/V = Q / (Qd/(A\epsilon_0)) = \frac{\epsilon_0 A}{d}$.
Solution: $C = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 25 \times 10^{-4}}{2 \times 10^{-3}} = 11.06 \times 10^{-12} \text{ F} = 11.06 \text{ pF}$.
Solution: The capacitance increases by a factor of $K$ (dielectric constant), i.e., $C' = KC$.
Solution: (i) It increases the capacitance. (ii) It prevents plates from sticking together. (iii) It increases the maximum operating voltage by reducing the electric field inside.
Solution: New capacitance $C' = K \times C = 4 \times 50 \text{ pF} = 200 \text{ pF}$.
Solution: Since battery is removed, charge $Q$ is constant. $Q = CV = 50 \times 200 = 10000 \text{ pC}$. New potential $V' = Q/C' = V/K = 200 / 4 = 50 \text{ V}$.
Solution: Potential $V_1 = \frac{k(4\pi R^2 \sigma)}{R} \propto R$. Potential $V_2 = \frac{k(4\pi(2R)^2 \sigma)}{2R} \propto 2R$. Since $V_2 > V_1$, positive charge will flow from the larger sphere to the smaller sphere.
Solution: In series, charge $Q$ is same. $V = V_1 + V_2 + V_3$. $\Rightarrow \frac{Q}{C_s} = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3}$. Thus, $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.
Solution: In parallel, potential $V$ is same. Total charge $Q = Q_1 + Q_2 + Q_3$. $\Rightarrow C_p V = C_1 V + C_2 V + C_3 V$. Thus, $C_p = C_1 + C_2 + C_3$.
Solution: (i) Series: $C_s = \frac{C_1 C_2}{C_1+C_2} = \frac{5 \times 10}{15} = 3.33 \mu\text{F}$. (ii) Parallel: $C_p = 5 + 10 = 15 \mu\text{F}$.
Solution: $1/C_s = 1/9 + 1/9 + 1/9 = 3/9 = 1/3$. Therefore, $C_s = 3 \text{ pF}$.
Solution: Since they are identical and in series, the voltage divides equally. $V_1 = V_2 = V_3 = 120/3 = 40 \text{ V}$.
Solution: In parallel, $V=50\text{V}$ for both. $Q_1 = C_1 V = 20\mu\text{F} \times 50\text{V} = 1000 \mu\text{C}$. $Q_2 = C_2 V = 30\mu\text{F} \times 50\text{V} = 1500 \mu\text{C}$.
Solution: Work done to add charge $dq$ at potential $V'$ is $dW = V' dq = (q/C) dq$. Total work $U = \int_{0}^{Q} \frac{q}{C} dq = \frac{1}{C} \left[\frac{q^2}{2}\right]_{0}^{Q} = \frac{Q^2}{2C}$. Using $Q=CV$, $U = \frac{1}{2}CV^2$.
Solution: $U = \frac{1}{2}CV^2 = \frac{1}{2} \times 12 \times 10^{-12} \times (50)^2 = 6 \times 10^{-12} \times 2500 = 1.5 \times 10^{-8} \text{ Joules}$.
Solution: $U = \frac{1}{2}CV^2$. We know $C = \epsilon_0 A/d$ and $V = Ed$. $U = \frac{1}{2} (\frac{\epsilon_0 A}{d}) (E^2 d^2) = \frac{1}{2} \epsilon_0 E^2 (Ad)$. Energy density $u = U / \text{volume} = U / (Ad) = \frac{1}{2}\epsilon_0 E^2$.
Solution: Decrease. Since battery is removed, charge $Q$ is constant. Initial energy $U_i = Q^2/2C$. After dielectric insertion, $C' = KC$. Final energy $U_f = Q^2/2(KC) = U_i / K$. Since $K>1$, $U_f < U_i$. (The missing energy is used in pulling the dielectric into the capacitor).
