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Worksheets3Q_GEN PHYSICS 2_SUMMATIVE TEST
Total questions: 50
Worksheet time: 25mins
A student performs two activities: Activity 1: A plastic comb is rubbed with dry hair. Activity 2: A charged rod is brought near a neutral metal sphere without touching it. Which statement describes what happens in each activity?
In both activities, objects must touch to charge.
In both activities, protons move between the objects.
In Activity 1, electrons transfer; in Activity 2, charges rearrange.
In Activity 1, charges rearrange; in Activity 2, electrons transfer.
After rubbing a balloon on dry hair, the balloon sticks to a wall. Explain why the balloon becomes charged.
Electrons are transferred from the hair to the balloon.
Neutrons move between the balloon and the hair.
The balloon creates electrons through friction.
Protons move from the hair to the balloon.
A small positive test charge is placed near a charged object and begins to move. What does this observation best show about the region around the charged object?
The region has an electric field present.
The region contains electric potential only.
The region contains moving electric charges.
The region has equal positive and negative charges.
Several point charges are placed near a point in space. Each charge produces an electric field at that point. According to the principle of superposition, how is the net electric field at that point determined?
By adding the magnitudes of all electric fields
By considering only the strongest electric field
By averaging the electric fields from each charge
By adding all individual electric fields as vectors
A long wire has charge density λ=3.0×10−6C/m . A point is located at a distance r=0.20m from the wire. The permittivity of free space is ε0=8.85×10−12C2/(N⋅m2) . Which statement correctly describes the electric field at this point due to the wire?
The field magnitude is 2.7×105N/C , it points toward the wire, and decreases with distance.
The field magnitude is 2.7×105N/C , it points away from the wire, and decreases with distance.
The field magnitude is 2.7×105N/C , it points along the wire, and increases with distance.
The field magnitude is 2.7×105N/C , it circles around the wire, and stays constant with distance.
A teacher wants to demonstrate charging by induction using a metal sphere, a charged rod, and a grounding wire. Which procedure correctly demonstrates charging by induction?
Heat the metal sphere to release electric charges.
Rub the metal sphere using a dry insulating cloth.
Touch the charged rod directly to the metal sphere.
Bring the rod near, ground the sphere, then remove both.
Two point charges are placed 0.50 m apart: q1 = +2 μC and q2 = −3 μC. What is the magnitude of the electric force between them?
0.024 N
0.108 N
0.216 N
2.16 N
A positive test charge is placed at different points around a negative charge. At which point will the electric field be strongest?
Where the field lines curve
Where the charge is neutral
Closest to the negative charge
Farthest from the negative charge
A point charge of +3.0 μC is fixed in space. A point is located 0.20 m away from the charge. Using Coulomb’s law, what is the electric field at that point, including its direction?
6.74 × 10^6 N/C, directed towards the positive charge
1.35 × 10^6 N/C, directed towards the positive charge
6.74 × 10^5 N/C, directed away from the positive charge
1.35 × 10^5 N/C, directed away from the positive charge
A uniform electric field of E = 200 CN passes through a flat rectangular sheet with area A = 0.5 m2 . The sheet is tilted at an angle θ relative to the field. What is the electric flux through the sheet if the tilt angle is θ=600 ?
ΦE = 0 N·m^2/C
ΦE=50N⋅m2/C
ΦE=70.7N⋅m2/C
ΦE=100N⋅m2/C
A sphere of radius 0.10 m carries a total charge of 4.0 μC uniformly distributed throughout its volume. Calculate the magnitude of the electric field at a point 0.05 m from the center of the sphere (inside the sphere).
1.80×103 N/C
1.80×104 N/C
8.99×103 N/C
8.99 × 104 N/C
A charge of q = 5 μC is moved across a potential difference of 120 V. What is the work done on the charge?
Work = 6.0×10−4C
Work = −6.0×10−4J
Work = 6.0×10−4J
Work = −6.0×10−4C
A charge Q = 40.0×10−12 C is uniformly distributed on a plastic sphere of 0.005 m radius. Find the electric potential at its surface when k = 8.99×109 N·m^2/C^2.
0.719 V
719 V
71.9 V
7.19 V
A set of equipotential lines is drawn around a metal sphere. The equipotential surfaces are perfect concentric circles, with values: 80 V (closest to sphere), 60 V, 40 V, 20 V. The equipotential lines are closest near the sphere and farther apart as you move away. Based on the equipotential lines, what is the direction of the electric field around the sphere?
Along the equipotential lines
Away from the sphere
In random directions
Toward the sphere
The electric potential along x-axis is given by V(x)=8x−4 where V is in volts and x is in meters. Find the electric field in the region.
±8 N/C
−8 N/C
8 N/C
None of the above
The electric potential along x-axis is given by where V〈x〉=8x -4 is in volts and x in meters. what is the magnitude and direction of the electric field in this region?
Ex=−8 N/C, pointing in the negative x-direction
Ex=−8 N/C, pointing in the positive x-direction
Ex=8 N/C, pointing in the negative x-direction
Ex=8 N/C, pointing in the positive x-direction
A parallel-plate capacitor has plates of area A=0.02m2 separated by d=0.01m . If the plate area is doubled while keeping the separation and voltage the same, which of the following occurs?
Capacitance doubles, stored charge doubles
Capacitance halves, stored charge stays the same
Capacitance doubles, stored charge stays the same
Capacitance stays the same, stored charge doubles
Find the equivalent capacitance of the series when three capacitors C1=10μF , C2=10μF , and C3=12μF are connected.
28×10−6F
2.8×10−6F
0.028×10−6F
0.28×10−6F
A digital camera uses a capacitor system to store energy for its flash. Two capacitors with capacitances 2.0×10−6F and 3.0×10−6F are connected in series to a 100 V battery. What is the total charge of the system?
8.3×10−5C
8.3×105C
8.4×10−6C
8.4×10−4C
Two students rub different objects: • Student 1 rubs a plastic comb with dry hair. • Student 2 rubs a glass rod with silk. Both objects become charged. Which conclusion best explains this observation?
Protons move between the objects.
Rubbing always creates new charges.
Only insulators can be charged by rubbing.
Electron transfer depends on the materials involved.
During an induction experiment, a positively charged rod is brought near a neutral metal sphere. Electrons move toward the side near the rod. What does this observation show?
New electrons are created inside the conductor.
Charges transfer only through direct contact.
Charges redistribute due to an external field.
Protons move freely inside the conductor.
Three point charges lie on a straight line at positions
• q1=+4.0 μC at x=0.00 m
• q2=−2.0 μC at x=0.30 m
• q3=+3.0 μC at x=0.70 m
Explains the direction of the net force on q2.
The force from q1 is stronger than the force from q3.
The force from q3 cancels the force from q1.
Both forces act in the same direction.
Only the nearest charge affects q2.
Two regions have electric fields of different strengths. Which conclusion is most reasonable?
The stronger field exerts a greater force on a charge.
The weaker field contains more electric charges.
Both fields exert the same force on all charges.
Electric fields depend only on electric potential.
Two positive point charges lie on the x-axis:
• Q1=+4.0 μC at x=0.00 m
• Q2=+3.0 μC at x=0.40 m
Point P is located at the midpoint, x=0.20 m. The electric field at P due to each charge is calculated, and the net electric field is found to be directed to the right. Which statement best explains why the net electric field points to the right?
Both electric fields point in the same direction.
The midpoint makes one electric field disappear.
The electric field from Q1 is stronger than from Q2.
The electric field from Q2 is stronger than from Q1.
A charge Q = 40.0×10−12 C is uniformly distributed on a plastic sphere of 0.005 m radius when k = 8.99×109N⋅m2/C2. Explains the effect of this potential on a positive test charge near the sphere.
The test charge would move toward the sphere spontaneously, regardless of field.
The test charge would gain potential energy as it moves away from the sphere.
The test charge would lose potential energy as it moves away from the sphere.
The test charge’s potential energy remains constant regardless of distance.
You observe an oddly shaped metal object with equipotential lines very close together near a sharp tip and widely spaced in smoother regions of the object. Based on the equipotential lines, what can be inferred about the electric field around the object?
The electric field is strongest at the sharp tip and weakest on smooth areas.
The electric field is strongest on smooth areas and weakest at sharp tips.
The electric field has the same strength everywhere.
There is no electric field because the object is metal.
The electric potential in a region is given by V(x,y)=4x+2y and the electric field is related to the potential by E=−(∂x∂Vi^+∂y∂Vj^) . Which statement correctly describes the electric field vector in this region?
The electric field is −4i^−2j^ N/C, pointing toward lower potential in both x and y directions.
The electric field is 4i^+2j^ N/C, pointing toward higher potential in both x and y directions.
The electric field is −2i^−4j^ N/C, pointing toward higher potential in both directions.
The electric field is −2i^−4j^ N/C, pointing toward lower potential in both directions.
The electric potential in a region is V(x)=3x2−6x and the electric field is related to the potential by Ex=−dxdV . At x = 1.0 m, which statement correctly describes the electric field?
The electric field is positive because the potential increases with x.
The electric field is negative because the potential decreases with x.
The electric field is zero because the potential is minimum at this point.
The electric field is zero because the potential is maximum at this point.
A cylindrical capacitor has a length L and radius R. If the distance between the cylinders is increased, which statement explains the capacitor’s behavior?
Capacitance stays the same and stored charge decreases.
A higher voltage is needed to store the same charge.
Capacitance increases and stored charge increases.
Capacitance increases but voltage stays the same.
A technician is checking a circuit board where three capacitors (10 µF, 10 µF, and 12 µF) are connected in parallel to improve energy storage. Which statement correctly explains the equivalent capacitance of the circuit?
The equivalent capacitance is 3.2 µF because parallel capacitors must be divided by the number of capacitors.
The equivalent capacitance is 10 µF because parallel connections keep capacitance the same.
The equivalent capacitance is 32 µF because all capacitances in parallel are added directly.
The equivalent capacitance is 12 µF because only the largest capacitor affects the total.
A student claims: “The net electric force on a charge is always zero if there are charges on both sides.” Which evaluation is most accurate?
The claim is incorrect because force varies.
The claim is correct because forces cancel.
The claim is correct for opposite charges.
The claim is correct only in conductors.
A simplified one-dimensional model of a lithium atom is shown:
· A nucleus with the charge q1=+3.0e at x=0.00m ;
· Two electrons, each with charge q2=q3=−1.6×10−19C , are located at x=1.0×10−10m and x=−1.0×10−10m ; and
· Point P is at x=2.0×10−10m .
After calculating the electric field at point P, a student concludes that the net electric field points to the left because the electrons produce a stronger electric field than the nucleus. Evaluate the student’s conclusion. Explain whether it is correct or incorrect and justify your reasoning.
Correct, because the electrons are closer to point P
Incorrect, because the nucleus has a larger total charge
Correct, because negative charges always dominate electric fields
Incorrect, because electric fields from electrons cancel each other
A uniform electric field of E=150N/C passes through a square sheet with area A=0.4m2 . Four students calculated the flux when the sheet is tilted at θ=40∘ :
• Student 1: ΦE=60N⋅m2/C
• Student 2: ΦE=50N⋅m2/C
• Student 3: ΦE=80N⋅m2/C
• Student 4: ΦE=42.4N⋅m2/C
Which student’s answer is correct?
Student 1
Student 2
Student 3
None of the above
A large, flat, non-conducting plate has a uniform surface charge density of 8.0μC/m2 . Four students calculate the electric field near the plate: . Who correctly calculated the electric field?
Albert
Alejandro
Alex
Allan
Which statement CORRECTLY interpret the potential energy change?
Andrea
Anne
Amber
All the above
A metal sphere has a charge of 6.0×10−12C and a radius of 0.010m . The electric potential at the surface of a charged sphere is given by V=rkQ where k=8.99×109Nm2/C2 . Which conclusion BEST explains the electric potential at the surface and how the sphere’s charge affects it?
The electric potential is positive because the sphere has a positive charge
The electric potential is negative because charge spreads on the surface
The electric potential is zero because the sphere is made of metal
The electric potential depends only on the size of the sphere
Between two nearby plates, the equipotential lines appear as evenly spaced vertical lines labeled 100 V, 90 V, 80 V, 70 V, …, decreasing toward one plate. Which conclusion BEST describes the electric field between the plates?
The field is strongest near the higher-potential plate and weakest near the lower-potential plate
The field is uniform and points from the higher-potential plate to the lower-potential plate
There is no electric field because the potentials are evenly spaced
The field changes direction between the plates
A spherical capacitor has its radius doubled and a dielectric slab is added between the spheres. Which conclusion best describes the effects on the capacitance?
Capacitance remains unchanged
Capacitance increases due to larger radius and added dielectric
Capacitance decreases due to larger radius and added dielectric
Capacitance increases due to larger radius but decreases due to dielectric
A digital camera uses a capacitor system for its flash. Two capacitors, C1=2.0×10−6F and C2=3.0×10−6F , are connected in series to a 100V battery. C1 has 41.50V while C2 has 27.67V . Which statement correctly explains why one capacitor has a larger voltage than the other in this series connection based on the relationship between capacitance, charge, and voltage?
The capacitor with smaller capacitance has a larger voltage drop because both capacitors store the same charge
The capacitor with larger capacitance has a larger voltage drop because both capacitors store the same charge
Both capacitors have the same voltage because the battery divides it equally
Voltage distribution depends on the battery type, not capacitance
Two point charges lie on the x-axis:
Q1 = +4.0 μC at x = 0.00 m;
Q2 = +3.0 μC at x = 0.40 m;
a test point P is at x = 0.20 m.
Which method correctly finds the net electric field at P?
Find the field from only one charge, then ignore the effect of the other charge.
Find the field from each charge, ignore the direction, then add the magnitudes.
Find the field from each charge, determine the correct direction for each, then add them.
Find the field from each charge, use the midpoint as the distance, then add the magnitudes.
You are designing a solar panel that acts like a flat sheet in a uniform vertical electric field E = 100 N/C. The panel can tilt to maximize or minimize the flux. Which orientation would maximize the electric flux, and how would you calculate it?
Tilt the panel at 90°; calculate ΦE = EA cos 90°.
Tilt the panel at 45°; calculate ΦE = EA cos 45°.
Tilt the panel at 60°; calculate ΦE = EA cos 60°.
Tilt the panel at 0°; calculate ΦE = EA cos 0°.
The electric potential in a region is V(x)=3x2−6x . A student wants to place a charged particle at a point where it experiences no electric force. Which electric field expression and position should the student construct and choose to satisfy this condition?
Ex = 6x − 6, place the particle at x = 1.0 m
Ex = −6x − 6, place the particle at x = 1.0 m
Ex = −6x + 6, place the particle at x = 1.0 m
Ex = −3x + 6, place the particle at x = 2.0 m
A camera flash circuit must store about 20 μF of capacitance. You are given three capacitors with values 10 μF, 10 μF, and 12 μF. You may connect them in parallel only; not all capacitors must be used. Which design choice best meets the circuit requirement without exceeding 22 μF?
Connect the 10 μF and 10 μF capacitors in parallel
Connect the 10 μF and 12 μF capacitors in parallel
Connect all three capacitors in parallel
Use only the 12 μF capacitor
A camera flash circuit must store about 20 μF of capacitance. You are given three capacitors with values 10 μF, 10 μF, and 12 μF. You may connect them in parallel only, but not all capacitors must be used., what will you do to make the camera store more energy?
Replace them with smaller capacitors in series
Remove one capacitor from the series
Connect the capacitors in parallel
Reduce the battery voltage
Two regions have different electric field strengths. Which conclusion is most reasonable?
The stronger field exerts a greater force on a charge.
The weaker field contains more charges.
Both fields exert the same force on all charges.
Electric fields depend only on electric potential.
Two students rub different objects: a plastic comb with hair and a glass rod with silk. Both become charged. Which conclusion best explains this?
Electrons moved from one material to another, so the objects gained opposite charges.
Protons moved from one material to another, so the objects gained the same charge.
Heat from rubbing created new charges inside each object.
Rubbing only makes objects dirty; the charge is just dust sticking to them.
Equipotential lines around a metal sphere are concentric circles, closest near the sphere. What is the direction of the electric field?
Toward the sphere.
Away from the sphere.
Along the equipotential lines.
In random directions.
A metal object has equipotential lines very close near a sharp tip and widely spaced elsewhere. What can you infer about the electric field?
It is strongest at the sharp tip and weakest on smooth areas.
It is strongest on smooth areas and weakest at the tip.
It has the same strength everywhere.
There is no electric field because it is metal.
Between two nearby plates, equipotential lines are evenly spaced at 100 V , 90 V , 80 V … decreasing toward one plate. What best describes the electric field?
The field is uniform and points from the higher-potential to lower-potential plate.
The field is strongest near the higher-potential plate and weakest near the lower.
There is no electric field because potentials are evenly spaced.
The field changes direction between the plates.
Who correctly calculated the work done by the field?
Andrea
Anne
Amber
None
