Electrostatic MDCAT MCQs

Explanation:

We are asked to find the radius of a sphere that has the same capacitance as a parallel plate Capacitor using mica as the dielectric.

Capacitance depends on geometry and dielectric material. For a parallel plate Capacitor: C = (ε₀ ε_r A) / d, where ε_r is the dielectric constant. For a spherical Capacitor: C = 4 π ε₀ R. By equating these, the radius of the equivalent sphere can be determined.

The method involves converting area to m², distance to meters, calculating the parallel plate capacitance, and setting it equal to the spherical Capacitor’s formula to solve for R. This ensures the two configurations store the same charge at the same potential.

Analogy: Think of two water tanks — one flat and one spherical — that store the same volume; the spherical tank’s radius must adjust to hold the same amount.

In summary, the radius can be determined by using the dielectric-adjusted parallel plate capacitance and equating it to the spherical Capacitor formula.

Explanation:

This problem requires designing a Capacitor Network to achieve a target capacitance and voltage rating using standard units.

Capacitors in series increase voltage tolerance but reduce equivalent capacitance. Parallel connections increase total capacitance while keeping voltage the same. A combination of series and parallel arrangements is needed to meet both requirements.

The approach: first calculate the number of Capacitors needed in series to safely handle 1000 V. Then, determine how many such series strings must be connected in parallel to reach 16 µF total capacitance. Multiplying series and parallel counts gives the total number of Capacitors required.

Analogy: Like stacking battery packs — series stacks increase voltage, parallel stacks increase capacity. The correct arrangement balances both requirements efficiently.

In summary, series and parallel combinations allow building a safe, optimal Network to achieve both voltage and capacitance requirements.

Explanation:

This involves charge redistribution between two spheres of different radii when connected.

Charge flows until both spheres reach the same potential, V = Q / (4 π ε₀ R). Total charge is conserved, but it distributes according to their radii. The larger sphere acquires proportionally more charge to equalize potentials.

Analogy: Two water tanks of different diameters connected by a pipe — water (charge) flows until the levels (potential) are equal.

In summary, connecting conductors leads to redistribution of charge proportional to their sizes to maintain equal potential.

Explanation:

We are examining a Capacitor behavior when disconnected and a dielectric is inserted.

Capacitance increases with the dielectric. Since the capacitor is isolated, the charge remains constant. Voltage across the plates decreases (V = Q / C) and stored energy (U = Q² / 2C) decreases.

Analogy: Stretching a spring under a fixed load; inserting a dielectric changes the potential while the charge (load) stays the same.

In summary, inserting a dielectric reduces voltage and stored energy while keeping the charge constant.

Explanation:

This problem connects capacitance to the geometry of a spherical capacitor.

For a spherical capacitor: C = 4 π ε₀ (R₁ R₂) / (R₂ – R₁), where R₁ and R₂ are inner and outer radii. Given C and the separation, we can solve for the outer radius R₂.

Analogy: Two nested spheres — the gap and inner radius determine how much charge the outer sphere can hold at a particular potential.

In summary, the outer radius can be determined from the capacitance and separation values.

Explanation:

We are asked to find the ratio of dielectric constants based on Electric Field reduction.

The field in a dielectric is E = E₀ / ε_r. Reduction of the field to 1/3 and 1/8 corresponds to the dielectric constants of polythene and PVC. The ratio PVC : polythene is obtained from these constants.

Analogy: Like insulating layers reducing Heat flow — stronger dielectrics reduce the field more.

In summary, dielectric constants are inversely proportional to the resulting Electric Field in the material.

Explanation:

Inserting a conductor between capacitor plates affects capacitance.

If the foil fully spans the area, it redistributes charges and can effectively short the capacitor. Capacitance can increase significantly or become extremely high depending on the placement.

Analogy: Like placing a conductive sheet between two charged surfaces, altering the Electric Field and stored charge capacity.

In summary, the capacitance increases significantly when a conductive foil is inserted.

Explanation:

We consider the combined effect of increasing plate separation and inserting a dielectric.

Capacitance: C = (ε₀ ε_r A) / d. Tripling the distance reduces C by 1/3; inserting the dielectric increases C by ε_r. Solving for ε_r gives the dielectric constant.

Analogy: Like stretching a spring reduces stiffness, inserting a medium increases the effective storage capacity.

In summary, the dielectric constant compensates for the increased separation to yield the observed capacitance.

Explanation:

Capacitance depends on series vs parallel arrangement.

Parallel: C_total = C₁ + C₂ + C₃ → maximum. Series: 1/C_total = 1/C₁ + 1/C₂ + 1/C₃ → minimum.

Analogy: Like resistors — series reduces total capacity, parallel adds it.

In summary, maximum occurs in parallel, minimum in series.

Explanation:

Stored energy is U = 1/2 C V². To maximize energy, maximize total capacitance.

Parallel arrangements increase total capacitance; series reduces it. For identical Capacitors at the same voltage, the combination with the largest effective C stores the most energy.

Analogy: Multiple water tanks — the more combined capacity, the more energy stored.

In summary, the arrangement with the largest equivalent capacitance at the same voltage stores the maximum energy.

Explanation:

This problem involves connecting two charged Capacitors and finding the resulting common voltage.

When Capacitors are connected in parallel, total charge is conserved. The combined voltage is calculated using V = Q_total / C_total, where Q_total is the sum of initial charges (Q = C × V). The final voltage is somewhere between the initial voltages, weighted by capacitance.

Analogy: Like mixing two water tanks of different volumes and water levels; the final level balances based on total volume.

In summary, the common voltage depends on total charge and combined capacitance.

Explanation:

This question asks about optimal dielectric properties for capacitor performance.

A good capacitor dielectric should have a high dielectric constant (k) to store more charge, and high dielectric strength (x) to withstand high voltage without breakdown. Materials with low k or low x are less efficient or unsafe for high voltage applications.

Analogy: Like a water tank with thick walls — you want both capacity and strength to hold more water safely.

In summary, ideal dielectric materials combine high k and high x for maximum energy storage and voltage tolerance.

Explanation:

This involves energy redistribution when a charged capacitor is connected to an uncharged one.

Total charge is conserved. After connection, charge redistributes proportionally to capacitances. The energy before and after connection is calculated using U = 1/2 C V². The difference between initial and final energy is the energy lost, usually as Heat in connecting wires or due to redistribution.

Analogy: Like pouring water from a smaller tank into a larger tank — some energy is lost to splashing or friction.

In summary, connecting capacitors leads to energy loss due to redistribution of charge.

Explanation:

When a disconnected capacitor’s plates are pulled apart, its capacitance decreases (C = εA/d), and charge remains constant.

Energy U = Q² / 2C. As C decreases, U actually increases because work is done against the attractive force between plates. The reasoning statement is correct — mechanical work must be applied — but the assertion about energy decreasing is false.

Analogy: Like stretching a spring that is isolated — energy increases due to work done, not decreases.

In summary, energy behavior depends on whether the battery is connected; mechanical work affects stored energy in disconnected capacitors.

Explanation:

When a capacitor is charged by a battery, part of the supplied energy is stored as electric potential energy in the capacitor.

The total work done by the battery is Q × V. The energy stored in the capacitor is U = 1/2 C V². Hence, only 50% of the supplied energy is stored; the other half is lost in the process, usually as Heat in the connecting wires or internal resistance.

Analogy: Filling a spring with energy — only part of your work is stored, some is dissipated.

In summary, a capacitor stores 50% of the battery’s supplied energy as potential energy.

Explanation:

For a capacitor connected to a battery, voltage is constant. Energy stored: U = 1/2 C V². Increasing plate separation reduces capacitance (C = εA/d).

Since V is constant, reducing C reduces U proportionally. Tripling d reduces C to 1/3, so stored energy reduces to 1/3.

Analogy: Like widening a spring’s attachment points while keeping the applied force constant — less energy is stored in the spring.

In summary, increasing plate separation with a constant voltage decreases stored energy proportionally.

Explanation:

Non-polar dielectrics do not have permanent dipole moments. Polar dielectrics have permanent dipoles that align in an Electric Field.

Among common materials: benzene is non-polar, whereas Alcohols, ammonia, and hydrochloric Acid are polar. Dielectric constant and orientation effects differ between polar and non-polar substances.

Analogy: Non-polar molecules are like small balls with no orientation; polar molecules are tiny magnets aligning with a field.

In summary, non-polar dielectrics do not exhibit permanent dipole alignment.

Explanation:

Capacitor combinations in series and parallel produce specific total capacitances. Series: 1/C_total = 1/C₁ + 1/C₂. Parallel: C_total = C₁ + C₂.

By testing combinations, values like 8 µF & 4 µF in series or parallel give the observed totals. The technique relies on systematic calculation using series/parallel formulas.

Analogy: Like mixing different water tanks — combined volumes depend on connection type.

In summary, specific total capacitances can be achieved by selecting suitable capacitor pairs and arranging them appropriately.

Explanation:

The stored energy in a dielectric material is given by U = 1/2 ε ε₀ E² × volume. Here, ε = dielectric constant, E = Electric Field.

Compute volume of the cube (side³), then multiply by dielectric-adjusted energy density. This gives the total energy stored inside the dielectric.

Analogy: Like a block of insulating material in a water flow — the material stores energy proportional to its volume and properties.

In summary, stored energy depends on dielectric constant, cube volume, and applied field.

Explanation:

A Van de Graaff generator produces high voltages by transferring charge to a large spherical conductor via a moving belt. It can accelerate charged particles in experiments.

Applications include nuclear Physics experiments, particle acceleration, and demonstrations of Electrostatics. The generator achieves high voltage with minimal current, making it safe for research setups.

Analogy: Like continuously filling a balloon with static charge to reach very high potential without bursting.

In summary, the Van de Graaff generator is primarily used to generate high voltage for Physics applications.

Explanation:

Capacitors can store and release energy in both AC and DC circuits. In AC, they allow Alternating Current to pass while blocking DC components. In DC, they can store energy, filter signals, or smooth voltage.

Analogy: Like a reservoir that can supply water intermittently or steady, depending on the system.

In summary, capacitors are versatile components used in both AC and DC circuits for energy storage and signal control.

Explanation:

Placing a grounded conductor near a charged object induces opposite charges on its near side due to electrostatic induction. The potential of the charged conductor decreases slightly, and its capacitance effectively increases.

Analogy: Like placing a neutral metal plate near a charged balloon — charges rearrange and the balloon’s field slightly weakens.

In summary, grounding allows induced charges to neutralize potential, altering the field and capacitance.

Explanation:

Capacitor plates store equal and opposite charges. Even if one plate is smaller, the charge on it equals the charge on the larger plate due to the conservation of charge in the capacitor system. Electric Field lines adjust to the geometry.

Analogy: Like a see-saw — the torque balance (charge balance) depends on opposite sides being equal, not size.

In summary, the smaller plate carries the same charge magnitude as the larger plate.

Explanation:

Connecting a charged body to the Earth allows excess charge to flow until its potential equals that of the Earth (0 V). Positive charge moves to the ground to neutralize the potential difference.

Analogy: Like draining excess water from a tank until it reaches the level of a larger reservoir.

In summary, grounding neutralizes a body’s potential.

Explanation:

A capacitor blocks steady DC after initial charging because current cannot flow through the dielectric. The effective resistance in DC is theoretically infinite.

Analogy: Like a closed gate in a water channel — no water passes after the initial surge.

In summary, for DC, a fully charged capacitor behaves as an open circuit with infinite resistance.

Explanation:

In a uniform Electric Field, field lines are parallel and equally spaced, representing a constant magnitude and direction of the field. They are not perpendicular to each other.

Analogy: Like evenly spaced parallel tracks on a railroad, indicating uniformity.

In summary, uniform fields are visualized as parallel, equally spaced lines.

Explanation:

Capacitance is defined as the charge stored per unit potential difference: C = Q / V. Its SI unit is the Farad (F), equivalent to coulomb/volt.

Analogy: Like measuring water held per meter of pressure — Farad is the electrical equivalent.

In summary, the Farad quantifies a capacitor’s ability to store charge per volt.

Explanation:

Connecting capacitors in parallel increases total capacitance while keeping voltage constant. This arrangement allows more charge to be stored without changing the applied potential.

Analogy: Like adding more buckets in parallel to store water while keeping water level unchanged.

In summary, parallel configuration enhances charge storage at constant voltage.

Explanation:

The ratio of charge to potential increase defines capacitance: C = Q / V. It is a measure of how much charge a conductor can hold per volt.

Analogy: Like the capacity of a bucket to hold water per unit height of water.

In summary, capacitance quantifies the charge storage capability relative to potential.

Explanation:

The presence of a charged conductor induces opposite charges on the nearby side of the uncharged conductor. The potential and charge distribution of the system changes, though total charge remains constant.

Analogy: Like placing a neutral balloon near a charged balloon — charges rearrange due to induction.

In summary, proximity to a charged conductor induces changes in charge and potential of the uncharged conductor.

Explanation:

The dielectric constant (k) of a material is the ratio of charge stored with the dielectric to charge stored in air under the same voltage: k = Q_dielectric / Q_air.

This shows how much more charge a capacitor can store when a dielectric is inserted. Higher k materials allow greater energy storage.

Analogy: Like replacing a small bucket with a flexible one — it can hold more water under the same pressure.

In summary, dielectric constant quantifies the enhancement in charge storage due to the dielectric material.

Explanation:

The electrostatic force between capacitor plates is F = 1/2 (Q² / εA) or equivalently F = 1/2 ε A E², where E = V/d. Here, ε is permittivity, A is plate area, and d is separation.

By calculating E from V/d and using plate area, the attractive force can be determined. This force arises due to opposite charges on the plates attracting each other.

Analogy: Like two magnets facing opposite poles — they pull toward each other with a calculable force.

In summary, plate separation, voltage, and area determine the electrostatic attraction between capacitor plates.

Explanation:

Potential difference across a capacitor is given by V = Q / C. With known charge and capacitance, V can be computed directly.

Analogy: Like measuring water height in a tank — the volume of water (charge) over the tank’s capacity (capacitance) gives the level (voltage).

In summary, voltage across a capacitor is directly proportional to stored charge and inversely proportional to capacitance.

Explanation:

When drops combine, volume adds up: V_total = 64 × V_small = 64 × (4/3 π r³). Radius of the larger drop: R = (64)^1/3 × r = 4r.

Charge is proportional to radius: Q ∝ R. Hence, total charge on larger drop scales with the number of combined drops’ volume and radius.

Analogy: Like merging small balloons into a big one — total air volume combines, and surface charge distributes over the larger radius.

In summary, combined drops result in a larger radius and correspondingly larger total charge.

Explanation:

When conductors are connected, total charge redistributes according to capacitance: Q = C × V. Charges do not necessarily equalize but distribute proportionally to capacitances (capacitors in parallel analogy).

Analogy: Like pouring water between two tanks of different sizes — final water level is proportional to tank volumes.

In summary, charge sharing depends on capacitances, and equalization occurs only if capacitances are identical.

Explanation:

For a capacitor, voltage is V = Q / C. Increasing charge Q while keeping C constant increases potential difference V. Capacitance remains unchanged as it depends on geometry and dielectric.

Analogy: Like adding more water to a fixed-size tank — water height (voltage) rises.

In summary, increasing stored charge increases voltage across the capacitor.

Explanation:

By definition, capacitance C = Q / V. It indicates how much charge a conductor can store per unit potential.

Analogy: Like a tank’s capacity for water per unit height — bigger tanks hold more water for same height.

In summary, capacitance measures a body’s ability to store charge for a given voltage.

Explanation:

When a charged capacitor discharges through a resistor, energy stored U = 1/2 C V² converts to Heat in the resistor. The resistor dissipates this energy as Joule heating.

Analogy: Like draining water through a narrow pipe — the energy is dissipated as frictional Heat.

In summary, the Heat released equals the initial energy stored in the capacitor during discharge.

Explanation:

Energy stored: U = 1/2 C V². Convert units (μF to F, kV to V) before calculation. This formula calculates electric potential energy stored between plates.

Analogy: Like compressing a spring — energy is proportional to the square of displacement (voltage).

In summary, capacitor energy depends on capacitance and square of applied voltage.

Explanation:

Inserting a dielectric reduces the electric field between plates: E = E₀ / k, because the dielectric polarizes and opposes the field. Voltage across a disconnected capacitor decreases accordingly.

Analogy: Like inserting an insulating sheet between charged plates — it reduces the effective field experienced.

In summary, the dielectric decreases the electric field strength inside a capacitor.

Explanation:

When a capacitor is disconnected from the battery, charge Q remains constant. Increasing plate separation reduces capacitance C = εA/d. Since Q is constant, voltage V = Q/C increases. Stored energy U = Q²/2C also increases because work is done to separate the plates.

Analogy: Like stretching an isolated spring — energy increases as you do work on it.

In summary, separating disconnected capacitor plates increases voltage and stored energy.

Explanation:

For a spherical conductor, the potential is the same throughout its volume. Therefore, the center has the same potential as the surface.

Analogy: Like water in a tank at rest — the pressure is uniform at all points at a given height.

In summary, the potential inside a conductor is equal to the surface potential.

Explanation:

When the battery is disconnected, charge remains constant. Increasing plate distance reduces capacitance C = εA/d, increasing voltage V = Q/C. Energy stored U = Q²/2C increases because mechanical work is done to separate the plates.

Analogy: Like pulling apart the plates of a spring-connected capacitor — energy rises with separation.

In summary, separating disconnected capacitor plates increases potential difference and stored energy.

Explanation:

Electric field inside a uniformly charged non-conducting sphere (r < R) increases linearly with distance: E ∝ r. Outside the sphere (r > R), it behaves as if all charge is at the center: E ∝ 1/r².

Analogy: Like water pressure in a spherical tank — increases inside and decreases outside.

In summary, field varies linearly inside and inversely with square outside the sphere.

Explanation:

When a dielectric is inserted into a charged capacitor, it polarizes and partially cancels the original field. For a disconnected capacitor, the electric field decreases: E = E₀/k. Voltage and energy stored adjust accordingly.

Analogy: Like placing an insulating material between two magnets — field strength reduces.

In summary, inserting a dielectric reduces the internal electric field of a charged capacitor.

Explanation:

Capacitors can function in both AC and DC circuits. In AC, they store and release energy, blocking DC flow. In DC, they charge and discharge, store energy, and filter voltage fluctuations.

Analogy: Like a reservoir supplying water intermittently in AC or storing water steadily in DC.

In summary, capacitors are used in both AC and DC circuits for energy storage and regulation.

Explanation:

With the battery connected, voltage is constant. Removing the dielectric reduces capacitance C = εA/d. To maintain voltage, extra charge flows from the battery to the capacitor to restore Q = C × V.

Analogy: Like removing a larger tank section — more water must be added to keep the level constant.

In summary, removing a dielectric in a battery-connected capacitor draws additional charge from the battery.

Explanation:

Charge has magnitude only; it does not have direction, making it a scalar. Its value does not depend on rotation or observer frame, so the reason correctly explains the assertion.

Analogy: Like counting coins — the total number is independent of orientation or perspective.

In summary, charge is scalar, invariant under rotation or change of reference frame.

Explanation:

Volume adds: R_big = (512)^1/3 r = 8r. Surface charge density σ = Q / (4πR²). Total charge adds linearly, so σ_big / σ_small = (512 × q) / (4π (8r)²) ÷ (q / 4π r²) = 8.

Analogy: Like merging small water droplets into a large one — surface properties scale with radius.

In summary, surface charge density of the larger drop is higher by a factor related to the cube root of number of combined drops.

Explanation:

The central charge experiences forces from the two outer charges along the Y-axis. These forces cancel along Y-axis if displaced slightly in that direction, but forces along other directions do not provide stable equilibrium. Stability is therefore along Y-axis only.

Analogy: Like a bead in a symmetric groove — stable in one direction, unstable in others.

In summary, equilibrium of central charge is conditionally stable along Y-axis displacement only.

Explanation:

The force on a charge due to multiple charges is the Vector sum of individual Coulomb forces: F = k Q₁ Q₂ / r². For corner B, consider contributions from adjacent sides and diagonal. Use Pythagoras to resolve diagonal force components.

Analogy: Like multiple ropes pulling a weight from different directions — the NET force is Vector sum.

In summary, NET force is determined by summing all Vector components of Coulomb forces from surrounding charges.

Explanation:

Electrostatic equilibrium occurs when NET force on each charge is zero. For like charges at triangle vertices, symmetry cannot satisfy equilibrium because repulsion cannot cancel completely unless charges vary in magnitude or sign.

Analogy: Like three magnets at triangle corners — all push away; perfect balance is impossible.

In summary, electrostatic equilibrium is not possible for identical like charges at equilateral triangle vertices.

Explanation:

Electric field from a point charge follows E = k Q / r². Doubling distance (r → 2r) reduces field by factor 4: E_new = E / 4.

Analogy: Like spreading water from a point source — intensity decreases with square of distance.

In summary, electric field diminishes with square of distance from the charge.

Explanation:

Electric field near a uniformly charged flat plate: E = σ / 2ε₀. Force on a test charge: F = q E. Substitute σ = 6 μC/m², q = 3 μC to determine F.

Analogy: Like a charged sheet attracting a small metal ball — force proportional to surface charge and test charge.

In summary, force is calculated using field of charged plate and test charge.

Explanation:

Although charge is quantized at the microscopic level, 1 coulomb contains ~10¹⁹ electrons. At macroscopic scales, individual charges are imperceptible, so charge distribution seems continuous.

Analogy: Like sand grains in a sandbox — individual grains exist, but collectively it appears smooth.

In summary, charge appears continuous macroscopically due to extremely large number of discrete electrons.