Electrostatics Quiz on Electric Potential

Electric field lines start from positive charges and end at negative charges, they are tangential to the electric field vector, denser in stronger fields, do not intersect, and their number is proportional to charge magnitude.

Applications include flash capacitors in digital cameras and use in heart defibrillators; a disadvantage is that they can cause unwanted electric shock after discharging.

The torque experienced is given by the expression $\tau = pE \sin{\theta}$, where $p$ is the dipole moment and $E$ is the electric field strength.

<p>The electric potential $V$ at a point due to a point charge $q$ is defined as $V = \frac{kq}{r}$, where $k$ is Coulomb's constant and $r$ is the distance from the charge.</p> Signup and view all the answers

<p>The total force acting on an electric dipole in a uniform electric field is zero, as the forces on the positive and negative charges cancel each other out.</p> Signup and view all the answers

<p>Electric field lines are denser in regions where the electric field has a larger magnitude and less dense in regions with smaller magnitude.</p> Signup and view all the answers

<p>The electric field vector at a point in space is tangential to the electric field line at that point, indicating the direction and strength of the electric force.</p> Signup and view all the answers

<p>Capacitors are used in the ignition system of automobile engines to eliminate sparking.</p> Signup and view all the answers

<p>Electrostatic induction is the phenomenon of charging without physical contact of the charged body.</p> Signup and view all the answers

<p>A dielectric is a non-conducting material with no free electrons; examples include ebonite, glass, and mica.</p> Signup and view all the answers

<p>Non-polar molecules have their positive and negative charge centers coinciding and have no permanent dipole moment, such as O2 and CO2.</p> Signup and view all the answers

<p>Electric polarization is the total dipole moment per unit volume of a dielectric.</p> Signup and view all the answers

<p>Dielectric breakdown occurs when a high external electric field tears atoms apart, allowing bound charges to become free charges and the dielectric to conduct electricity.</p> Signup and view all the answers

<p>Electric field lines do not intersect because if they did, a charge placed at the intersection would experience two different electric field vectors, which is physically impossible.</p> Signup and view all the answers

<p>Quantization of charges means that the charge in an object can be expressed as $q = ne$, where $n$ is an integer and $e$ is the elementary charge.</p> Signup and view all the answers

<p>Being inside a car is safer because the metal body provides electrostatic shielding, and the electric field inside the car is zero during lightning.</p> Signup and view all the answers

<p>CJ represents the length of the potentiometer wire where the potential difference is equal to the emf of the cell, resulting in zero current through the galvanometer.</p> Signup and view all the answers

<p>The emf of a cell is directly proportional to the balancing length, expressed as $\epsilon \propto l$.</p> Signup and view all the answers

<p>The primary circuit consists of the battery, key, and the potentiometer wire.</p> Signup and view all the answers

<p>The galvanometer detects the flow of current, indicating when the potential difference is balanced and showing zero deflection.</p> Signup and view all the answers

<p>The formula is $\epsilon = I \cdot r \cdot l$.</p> Signup and view all the answers

<p>The first cell's emf is measured to find a balancing length, then the second cell's emf is measured similarly for comparison.</p> Signup and view all the answers

<p>No current flows through the galvanometer, resulting in zero deflection.</p> Signup and view all the answers

<p>The two circuits are the primary circuit and the secondary circuit.</p> Signup and view all the answers

<p>Kirchhoff's current rule states that the sum of currents entering a junction equals the sum of currents leaving it, expressed as 𝐼1 − 𝐼𝐺 − 𝐼3 = 0 at junction B, and 𝐼2 + 𝐼𝐺 − 𝐼4 = 0 at junction D.</p> Signup and view all the answers

<p>By substituting 𝐼𝐺 = 0 into equations (1), (2), (3), and (4), we find that 𝐼1 = 𝐼3 and 𝐼2 = 𝐼4.</p> Signup and view all the answers

<p>Zero deflection of the galvanometer indicates a balanced condition where the ratios of resistances in the bridge are equal, allowing us to determine the unknown resistance.</p> Signup and view all the answers

<p>The relationship is given by $rac{P}{Q} = rac{l_1}{l_2}$, where $l_1$ and $l_2$ are the lengths of the wire on either side of the jockey.</p> Signup and view all the answers

<p>Resistance values R and S correspond to the lengths AJ ($l_1$) and JB ($l_2$) of the bridge wire.</p> Signup and view all the answers

<p>Interchanging P and Q minimizes the effect of any end resistance and allows for the average value of P to be calculated more accurately.</p> Signup and view all the answers

<p>The relationship is expressed as 𝐼1 𝑃 + 𝐼𝐺 𝐺 − I2 𝑅 = 0, indicating that the sum of voltage drops equals the sum of voltage rises.</p> Signup and view all the answers

<p>Manganin is typically used for the bridge wire due to its uniform resistivity and low temperature coefficient of resistance.</p> Signup and view all the answers

<p>The expression is $V_1 = \frac{q}{4\pi\epsilon_0 r_1}$.</p> Signup and view all the answers

<p>The expression is $V_2 = -\frac{q}{4\pi\epsilon_0 r_2}$.</p> Signup and view all the answers

<p>The electric potential is expressed as $V = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$.</p> Signup and view all the answers

<p>The simplification involves substituting $\frac{1}{r_1} = \frac{1}{r} + \frac{a \cos \theta}{r}$ and $\frac{1}{r_2} = \frac{1}{r} - \frac{a \cos \theta}{r}$.</p> Signup and view all the answers

<p>When $ heta = 0$, $V = \frac{P \cos \theta}{4\pi\epsilon_0 r^2}$.</p> Signup and view all the answers

<p>When $ heta = 180$, $V = -\frac{P \cos \theta}{4\pi\epsilon_0 r^2}$.</p> Signup and view all the answers

<p>When $ heta = 90$, $V = 0$.</p> Signup and view all the answers

<p>The relationship is given by $V = \frac{2qa \cos \theta}{4\pi\epsilon_0 r^2}$.</p> Signup and view all the answers