Physics: Circuits and Resistance Concepts

<p>The resistance of a wire decreases as its cross-sectional area increases. (B)</p> Signup and view all the answers

<p>9.0 x 108 N/C (A)</p> Signup and view all the answers

<p>2 A (A)</p> Signup and view all the answers

<p>2.4 x 107 N (A)</p> Signup and view all the answers

<p>2 A (A)</p> Signup and view all the answers

<p>The electric flux will increase. (C)</p> Signup and view all the answers

<p>20 C (D)</p> Signup and view all the answers

<p>It is negative because the toner particle has a negative charge. (C)</p> Signup and view all the answers

<p>12 Nm²/C (C)</p> Signup and view all the answers

<p>20 C (D)</p> Signup and view all the answers

<p>10 Nm²/C (C)</p> Signup and view all the answers

<p>The total charge is doubled. (D)</p> Signup and view all the answers

<p>The electric flux increases. (C)</p> Signup and view all the answers

<p>0.5 A (C)</p> Signup and view all the answers

<p>1.6 x 10^7 m (B)</p> Signup and view all the answers

<p>The current is directly proportional to the voltage and inversely proportional to the resistance. (D)</p> Signup and view all the answers

<p>1.8×10^4 N/C (C)</p> Signup and view all the answers

<p>The conductor with 0.25 A for 5 hours transfers more charge. (C)</p> Signup and view all the answers

<p>Add the magnitudes of the forces due to q1 and q2 directly without considering their directions. (B)</p> Signup and view all the answers

<p>2400 C (D)</p> Signup and view all the answers

<p>0.12 A (A)</p> Signup and view all the answers

<p>The electric potential decreases as the distance from the charge increases. (A)</p> Signup and view all the answers

<p>They filter out unwanted frequencies, allowing only the desired signal to pass. (D)</p> Signup and view all the answers

<p>The electric potential is zero. (D)</p> Signup and view all the answers

<p>Capacitors smooth out fluctuations in voltage by storing and releasing charge. (C)</p> Signup and view all the answers

<p>To control the rate at which the circuit charges and discharges, determining the timing cycle. (B)</p> Signup and view all the answers

<p>The electric potential energy is reduced by a factor of four. (D)</p> Signup and view all the answers

<p>To store and release electrical energy to smooth out fluctuations in voltage. (C)</p> Signup and view all the answers

<p>The electric field strength is reduced by a factor of four. (B)</p> Signup and view all the answers

<p>The electric field points radially outward from the charge. (C)</p> Signup and view all the answers

<p>The capacitance increases, the charge on the plates remains the same, and the electric field decreases. (D)</p> Signup and view all the answers

<p>The electric field is uniform and points from the positive plate to the negative plate. (B)</p> Signup and view all the answers

<p>Capacitors can release the stored energy quickly to generate a high-intensity flash. (C)</p> Signup and view all the answers

<p>The charge on the capacitor increases. (D)</p> Signup and view all the answers

<p>The charge on the capacitor increases. (D)</p> Signup and view all the answers

<p>The electric field strength decreases. (C)</p> Signup and view all the answers

<p>The total amount of charge passing through a conductor in a given time. (A)</p> Signup and view all the answers

<p>The electric flux remains the same because it depends only on the enclosed charge. (A)</p> Signup and view all the answers

<p>The capacitance increases to 10 µF, the charge increases to 100 µC, and the voltage remains at 10 V. (B)</p> Signup and view all the answers

<p>The presence of other charges nearby. (B)</p> Signup and view all the answers

<p>The capacitance increases to 40 µF, the charge remains at 60 µC, and the voltage decreases to 1.5 V. (D)</p> Signup and view all the answers

<p>When the surface is perpendicular to the electric field. (C)</p> Signup and view all the answers

<p>The electric flux becomes zero. (C)</p> Signup and view all the answers

<p>The capacitance increases but the charge does not change. (B)</p> Signup and view all the answers

Relationship between current, voltage, and resistance.

The force between two charges due to their electric field.

Rate of electric charge passing a point in a circuit.

Total electric charge moved over time in a circuit.

Defines the force between two point charges.

A field around charged objects affecting other charges.

Combining forces by considering their directions.

How different conductors respond to electric current.

Calculates net electric field from multiple charges by summing vector contributions.

Resistance increases as the length of a wire increases, due to longer pathways for current.

Resistance decreases as the cross-sectional area of a wire increases, providing more routes for current.

Use Coulomb’s law to find electric fields and vector addition to sum them at a point.

Resistance of a wire increases by a factor of 4 if length is doubled and area is halved, according to R=ρ L/A.

Current through a light bulb is calculated as I=V/R, where V=24V and R=12Ω gives 2A.

Finds the electrostatic force between two charges using Coulomb's law. Separation increases force's distance sensitivity.

In a circuit, resistance, voltage, and current are related by Ohm's law: V=IR.

The electric field passing through a surface, representing how much field penetrates that surface.

Occurs when a surface is perpendicular to the electric field lines.

Represents the total charge (Q) passing through a conductor in a specific time (t) given the current (I).

Electric flux remains the same when the radius of a spherical surface is changed, as it depends only on the enclosed charge.

When a dielectric is inserted in a charged capacitor, its capacitance increases, maintaining the same charge and voltage.

Strength of the electric field depends on the magnitude of charge and distance; the medium affects it too, but nearby charges do not affect it.

In a disconnected charged capacitor, inserting a dielectric increases capacitance but reduces voltage while charge stays the same.

Doubling the radius of a spherical surface around a point charge does not change the electric flux.

The ability of a capacitor to store charge per unit voltage.

The direction of an electric field created by a positive charge points radially outward.

When the distance from a point charge is doubled, the electric field strength is reduced by a factor of four.

Inserting a dielectric increases capacitance, decreases charge, and reduces electric field.

The electric field between two parallel plates is uniform and points from positive to negative.

Capacitors are used in flash units because they release stored energy quickly for a high-intensity flash.

Capacitance can increase while charge remains the same if voltage remains constant.

The charge on the plates of a capacitor can decrease when a dielectric is introduced with constant voltage.

The electric potential increases as the distance from the charge increases.

Capacitors filter out unwanted frequencies, allowing only the desired signal to pass.

The electric potential is zero at a point halfway between +Q and -Q.

Capacitors smooth out fluctuations in voltage by storing and releasing charge.

Capacitors control the rate of charging and discharging, determining the timing cycle.

If the distance between two charges is doubled, the electric potential energy is reduced by a factor of four.

Capacitors store and release electrical energy to smooth out fluctuations in voltage.

Electric potential energy relates inversely to the square of the distance between charges.

An imaginary closed surface surrounding a charge for analysis.

This equation calculates total electric charge transferred over time.

Total charge increases when current flows longer with constant current.

Electric flux is determined by the charge within the Gaussian surface, not surrounding charges.

The strength of the electric force per unit charge in a field.

Electric flux is maximum when surface is parallel to the electric field direction.

Total charge transferred is calculated as 15 C from Q=I × t using values provided.