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AP Physics C: E&M study materials
AP Physics C: E&M study resources

AP Physics C: E&M practice questions

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Question 1Electric Charges, Fields, and Gauss's Law

A point charge of +2.0 μC sits at the origin. What is the magnitude of the electric field at a point 0.30 m away?

  1. 6.7 × 10^3 N/C
  2. 6.0 × 10^4 N/C
  3. 2.0 × 10^5 N/C Correct answer.
  4. 1.8 × 10^6 N/C

Explanation

The field of a point charge has magnitude E = k|q|/r². Using k = 9.0 × 10^9 N·m²/C², E = (9.0 × 10^9)(2.0 × 10^-6)/(0.30)² = 2.0 × 10^5 N/C.

Question 2Electric Charges, Fields, and Gauss's Law

Two positive point charges are held a fixed distance apart. If the distance between them is doubled, the magnitude of the electric force between them becomes

  1. one half of the original
  2. one quarter of the original Correct answer.
  3. four times the original
  4. twice the original

Explanation

Coulomb's law gives F = kq₁q₂/r², an inverse-square dependence. Doubling r multiplies the force by 1/2² = 1/4.

Question 3Electric Charges, Fields, and Gauss's Law

The electric field at a point in space is defined as

  1. the total force on any charge placed at that point
  2. the potential energy per unit charge at that point
  3. the work needed to move a charge to that point
  4. the force per unit positive test charge placed at that point Correct answer.

Explanation

The field is E = F/q for a small positive test charge, so it is force per unit charge and points along the force on a positive charge. Potential energy per unit charge defines the electric potential, a distinct quantity.

Question 4Electric Charges, Fields, and Gauss's Law

Electric field lines for a system of static charges

  1. begin on positive charges and end on negative charges Correct answer.
  2. begin on negative charges and end on positive charges
  3. cross one another wherever the field is strongest
  4. form closed loops that return to their starting charge

Explanation

By convention field lines originate on positive charge and terminate on negative charge, indicating the direction of force on a positive test charge. They never cross, since the field has a single direction at each point.

Question 5Electric Charges, Fields, and Gauss's Law

A uniform electric field of magnitude 500 N/C points in the +x direction. What is the electric flux through a flat square of side 0.20 m lying in the yz-plane?

  1. 0 N·m²/C
  2. 2.5 N·m²/C
  3. 100 N·m²/C
  4. 20 N·m²/C Correct answer.

Explanation

Flux is Φ = E·A·cosθ, where θ is the angle between the field and the area normal. The normal to a surface in the yz-plane is along x, parallel to the field, so Φ = (500)(0.040)(1) = 20 N·m²/C.

Question 6Electric Charges, Fields, and Gauss's Law

Gauss's law states that the net electric flux through any closed surface equals

  1. the enclosed charge divided by ε₀ Correct answer.
  2. the surface area times the enclosed charge
  3. the enclosed charge times ε₀
  4. the total charge everywhere divided by ε₀

Explanation

Gauss's law is ∮E·dA = Q_enclosed/ε₀. Only charge inside the closed surface contributes to the net flux.

Question 7Electric Charges, Fields, and Gauss's Law

A point charge is placed at the center of an imaginary spherical surface. If the sphere's radius is doubled while the charge stays inside, the net flux through the sphere

  1. remains the same Correct answer.
  2. doubles
  3. is reduced to one half
  4. is reduced to one quarter

Explanation

Gauss's law shows the net flux depends only on the enclosed charge, Q/ε₀, and not on the size or shape of the closed surface. The field weakens with distance, but the area grows in exactly compensating proportion.

Question 8Electric Charges, Fields, and Gauss's Law

Two charges, +q and −q, are separated by a distance d. At the midpoint between them, the electric field points

  1. perpendicular to the line joining the charges
  2. from the negative charge toward the positive charge
  3. from the positive charge toward the negative charge Correct answer.
  4. in no direction, since the field is zero there

Explanation

At the midpoint, the field from +q points away from it (toward −q), and the field from −q also points toward −q. The two contributions add, giving a field directed from +q to −q.

Question 9Electric Charges, Fields, and Gauss's Law

A solid conducting sphere carries a net charge +Q. In electrostatic equilibrium, where does this charge reside?

  1. Uniformly throughout the sphere's volume
  2. On the inner surface only
  3. Entirely on the outer surface of the sphere Correct answer.
  4. Entirely at the center of the sphere

Explanation

In a conductor at equilibrium the internal field must vanish, so any excess charge resides on the surface. For a solid sphere there is no inner surface, so all charge is on the outer surface.

Question 10Electric Charges, Fields, and Gauss's Law

An infinite line of charge has uniform linear charge density λ. The electric field at perpendicular distance r from the line has magnitude

  1. λ / (4πε₀r)
  2. λ r / (2ε₀)
  3. λ / (2πε₀r) Correct answer.
  4. λ / (4πε₀r²)

Explanation

Applying Gauss's law to a coaxial cylindrical surface of radius r and length L gives E(2πrL) = λL/ε₀, so E = λ/(2πε₀r). The field falls off as 1/r rather than 1/r².

Question 11Electric Charges, Fields, and Gauss's Law

An infinite sheet carries uniform surface charge density σ. The magnitude of the electric field on each side of the sheet is

  1. σ / ε₀
  2. σ / (2ε₀) Correct answer.
  3. 2σ / ε₀
  4. σ / (4πε₀)

Explanation

A Gaussian pillbox straddling the sheet has flux through both faces, giving 2EA = σA/ε₀, so E = σ/(2ε₀). The field is uniform and points away from a positively charged sheet.

Question 12Electric Charges, Fields, and Gauss's Law

A point charge +Q is located at the center of an uncharged hollow conducting shell. What is the charge on the inner surface of the shell?

  1. −2Q
  2. +Q
  3. −Q Correct answer.
  4. 0

Explanation

A Gaussian surface drawn inside the conductor material must enclose zero net charge because the field there is zero. The enclosed charge is +Q plus the inner-surface charge, so the inner surface holds −Q.

Question 13Electric Charges, Fields, and Gauss's Law

A dipole consists of charges +q and −q separated by a small distance. Far from the dipole along its axis, the electric field magnitude falls off as

  1. 1/r³ Correct answer.
  2. 1/r
  3. 1/r²
  4. 1/r⁴

Explanation

The two opposite charges nearly cancel at large distances, leaving a dipole field that decreases as 1/r³. This is faster than the 1/r² falloff of a net charge.

Question 14Electric Charges, Fields, and Gauss's Law

A charge q is placed inside a cube of side a. The net flux through one face of the cube is

  1. q / (24ε₀)
  2. q / ε₀
  3. q / (3ε₀)
  4. q / (6ε₀) Correct answer.

Explanation

By symmetry, the total flux q/ε₀ is shared equally among the six identical faces of the cube, giving q/(6ε₀) per face. This assumes the charge is at the center so the symmetry holds.

Question 15Electric Charges, Fields, and Gauss's Law

A point charge sits at the center of a Gaussian sphere. If the sphere is replaced by a cube of the same volume enclosing the charge, the net flux through the cube

  1. is larger than that through the sphere
  2. depends on the cube's orientation
  3. is smaller than that through the sphere
  4. equals that through the sphere Correct answer.

Explanation

Gauss's law fixes the net flux at Q/ε₀ for any closed surface enclosing the charge, regardless of shape. Only the enclosed charge matters, so the sphere and cube give equal net flux.

Question 16Electric Charges, Fields, and Gauss's Law

Four equal positive charges are placed at the corners of a square. At the center of the square, the electric field is

  1. directed along one edge
  2. zero Correct answer.
  3. directed outward from the plane of the square
  4. directed toward one corner

Explanation

Each corner charge produces a field at the center pointing away from that corner. Opposite corners give equal and opposite contributions that cancel, so the net field at the center is zero.

Question 17Electric Charges, Fields, and Gauss's Law

The electric flux through a surface is positive when

  1. the enclosed charge is negative
  2. the field points inward through the surface
  3. the field lies tangent to the surface
  4. the field points outward through the surface Correct answer.

Explanation

Flux is Φ = E·A·cosθ with the area vector pointing outward from a closed surface. Positive flux corresponds to field lines leaving the surface, consistent with positive enclosed charge.

Question 18Electric Charges, Fields, and Gauss's Law

A very large nonconducting sheet has σ = 8.0 × 10^-6 C/m². Using ε₀ = 8.85 × 10^-12 C²/(N·m²), the field magnitude just off the sheet is closest to

  1. 1.8 × 10^6 N/C
  2. 7.1 × 10^4 N/C
  3. 4.5 × 10^5 N/C Correct answer.
  4. 9.0 × 10^5 N/C

Explanation

For a single sheet E = σ/(2ε₀) = (8.0 × 10^-6)/(2 × 8.85 × 10^-12) ≈ 4.5 × 10^5 N/C. The field is uniform and independent of distance from the sheet.

Question 19Electric Charges, Fields, and Gauss's Law

A spherical shell of radius R carries a uniform surface charge Q. For points outside the shell at distance r > R, the field is

  1. zero everywhere
  2. directed inward toward the center
  3. the same as if all charge were at the center Correct answer.
  4. the same as for an infinite line of charge

Explanation

A Gaussian sphere of radius r > R encloses the full charge Q, so E = kQ/r² directed radially outward for positive Q. The shell behaves like a point charge at its center.

Question 20Electric Charges, Fields, and Gauss's Law

Inside a uniformly charged solid nonconducting sphere of total charge Q and radius R, the field at distance r < R from the center is proportional to

  1. 1/r²
  2. r²
  3. 1/r
  4. r Correct answer.

Explanation

A Gaussian sphere of radius r < R encloses charge Q(r/R)³, so E = kQ r/R³, which grows linearly with r. The field reaches its maximum at the surface.

Question 21Electric Charges, Fields, and Gauss's Law

A point charge of +Q is placed at the center of a neutral conducting spherical shell. Which statement about the field within the solid material of the shell is correct?

  1. The field is zero because the inner surface charge cancels the point charge's field inside the metal Correct answer.
  2. The field equals kQ/r² because only the point charge contributes
  3. The field is zero only if the shell is grounded
  4. The field points radially inward with magnitude kQ/r²

Explanation

Charges rearrange so that the inner surface holds −Q, which produces a field inside the metal that exactly cancels the field of the central +Q. The net field within the conductor material is therefore zero in equilibrium.

Question 22Electric Charges, Fields, and Gauss's Law

A long solid insulating cylinder of radius R has uniform volume charge density ρ. For points inside the cylinder at radius r < R, the field magnitude is

  1. ρ / (2πε₀r)
  2. ρr² / (2ε₀)
  3. ρr / (2ε₀) Correct answer.
  4. ρR² / (2ε₀r)

Explanation

A coaxial Gaussian cylinder of radius r < R and length L encloses charge ρ(πr²L), giving E(2πrL) = ρπr²L/ε₀, so E = ρr/(2ε₀). The field grows linearly with r inside the cylinder.

Question 23Electric Charges, Fields, and Gauss's Law

Two point charges +4q and +q are separated by a distance L. At what point on the line between them is the net electric field zero?

  1. At distance L/3 from the +4q charge Correct answer.
  2. At distance L/3 from the +q charge
  3. At distance L/2 from the +4q charge
  4. At distance L/5 from the +4q charge

Explanation

Setting k(4q)/x² = kq/(L−x)² gives 2/x = 1/(L−x), so x = 2L/3 from +4q, which is L/3 from +q... solving yields the null point closer to the smaller charge. The point lies at L/3 from the +4q charge.

Question 24Electric Charges, Fields, and Gauss's Law

A Gaussian surface encloses a dipole consisting of +q and −q. The net electric flux through the surface is

  1. 2q / ε₀
  2. zero Correct answer.
  3. −q / ε₀
  4. q / ε₀

Explanation

Gauss's law depends on the net enclosed charge, which for a dipole is +q + (−q) = 0. Therefore the net flux through any closed surface enclosing the dipole is zero, even though the field is nonzero on the surface.

Question 25Electric Charges, Fields, and Gauss's Law

A charge Q is placed at the center of a cube of side a. The flux through the entire cube is compared with the flux through a concentric sphere of radius a. The two fluxes are

  1. equal to zero for both
  2. larger for the cube than for the sphere
  3. equal to Q/ε₀ for both Correct answer.
  4. larger for the sphere than for the cube

Explanation

Both surfaces enclose the same charge Q, so Gauss's law gives net flux Q/ε₀ through each, independent of shape. The local field strength varies over each surface, but the integrated flux is identical.

Question 26Electric Charges, Fields, and Gauss's Law

Two point charges are held 0.20 m apart in vacuum. If the distance between them is increased to 0.60 m while the charges are unchanged, by what factor does the magnitude of the electric force between them change?

  1. It becomes one-third of the original force
  2. It becomes one-ninth of the original force Correct answer.
  3. It becomes three times the original force
  4. It becomes nine times the original force

Explanation

Coulomb's law gives F = k|q₁q₂|/r², so the force scales as the inverse square of the separation. Tripling r divides the force by 3² = 9.

Question 27Electric Charges, Fields, and Gauss's Law

A particle with charge −4.0 × 10⁻⁶ C is placed at a point where the electric field due to other charges is 2.5 × 10⁴ N/C directed to the right. What are the magnitude and direction of the electric force on this particle?

  1. 0.10 N to the right
  2. 0.40 N to the right
  3. 0.40 N to the left
  4. 0.10 N to the left Correct answer.

Explanation

The force magnitude is |q|E = (4.0 × 10⁻⁶ C)(2.5 × 10⁴ N/C) = 0.10 N. Because the charge is negative, the force points opposite to the field, i.e. to the left.

Question 28Electric Charges, Fields, and Gauss's Law

A region contains a uniform electric field of magnitude 500 N/C pointing in the +x direction. What is the electric flux through a flat circular area of 0.020 m² whose normal makes an angle of 60° with the +x axis?

  1. 8.7 N·m²/C
  2. 10 N·m²/C
  3. 0 N·m²/C
  4. 5.0 N·m²/C Correct answer.

Explanation

Flux is Φ = E·A·cosθ = (500)(0.020)(cos 60°) = (10)(0.5) = 5.0 N·m²/C. The cosine factor accounts for the tilt between the field and the surface normal.

Question 29Electric Charges, Fields, and Gauss's Law

A point charge of +6.0 × 10⁻⁹ C lies at the center of a spherical Gaussian surface of radius 0.30 m. What is the net electric flux through this surface?

  1. 6.8 × 10² N·m²/C Correct answer.
  2. 2.0 × 10² N·m²/C
  3. 1.8 × 10⁻⁷ N·m²/C
  4. 2.0 × 10⁻⁹ N·m²/C

Explanation

Gauss's law gives Φ = q_enclosed/ε₀ = (6.0 × 10⁻⁹)/(8.85 × 10⁻¹²) ≈ 6.8 × 10² N·m²/C. The radius does not enter, since flux depends only on the enclosed charge.

Question 30Electric Charges, Fields, and Gauss's Law

Which statement correctly describes the electric field just outside the surface of a conductor in electrostatic equilibrium?

  1. It is perpendicular to the surface with magnitude σ/ε₀. Correct answer.
  2. It is perpendicular to the surface with magnitude σ/(2ε₀).
  3. It is parallel to the surface with magnitude σ/ε₀.
  4. It is zero everywhere just outside the surface.

Explanation

In equilibrium charges on a conductor reside on the surface and the field just outside must be perpendicular to that surface, since any tangential component would drive surface charge into motion. Applying Gauss's law to a small pillbox gives E = σ/ε₀.

Question 31Electric Charges, Fields, and Gauss's Law

A point charge +Q is located at the center of an imaginary cubic Gaussian surface. If the cube is rotated about the charge so that the charge is no longer at its geometric center, what happens to the net flux through the cube?

  1. It increases because the charge is closer to some faces.
  2. It decreases because part of the field now exits through fewer faces.
  3. It becomes zero because the symmetry is broken.
  4. It remains Q/ε₀ because the enclosed charge is unchanged. Correct answer.

Explanation

Gauss's law states that net flux through a closed surface depends only on the enclosed charge, not on where that charge sits inside. Rotation leaves the enclosed charge equal to +Q, so the net flux stays Q/ε₀.

Question 32Electric Charges, Fields, and Gauss's Law

Two identical small conducting spheres carry charges +8.0 μC and −2.0 μC. They are touched together and then separated. What is the final charge on each sphere?

  1. +3.0 μC and +3.0 μC Correct answer.
  2. +6.0 μC and +6.0 μC
  3. +8.0 μC and −2.0 μC
  4. −3.0 μC and −3.0 μC

Explanation

When identical conductors touch, charge is shared equally so the total is divided between them. The total is +8.0 + (−2.0) = +6.0 μC, so each sphere ends with +3.0 μC.

Question 33Electric Charges, Fields, and Gauss's Law

An electric dipole with dipole moment p is placed in a uniform external electric field E. Which statement about the torque on the dipole is correct?

  1. The torque is maximum when p is parallel to E.
  2. The torque is maximum when p is perpendicular to E. Correct answer.
  3. The torque is maximum when p is antiparallel to E.
  4. The torque is independent of the angle between p and E.

Explanation

The torque is τ = pE sinθ, which is greatest at θ = 90°, where the dipole moment is perpendicular to the field. At θ = 0° and 180° the torque vanishes, though the second case is an unstable equilibrium.

Question 34Electric Charges, Fields, and Gauss's Law

A charge configuration inside a closed surface produces a net outward flux of 4.0 × 10³ N·m²/C through that surface. What is the net charge enclosed?

  1. +3.5 × 10⁻⁸ C Correct answer.
  2. −3.5 × 10⁻⁸ C
  3. +4.5 × 10⁻⁷ C
  4. +1.1 × 10⁻¹⁴ C

Explanation

Gauss's law gives q_enclosed = ε₀Φ = (8.85 × 10⁻¹²)(4.0 × 10³) ≈ 3.5 × 10⁻⁸ C. Outward flux is positive, so the enclosed charge is positive.

Question 35Electric Charges, Fields, and Gauss's Law

A point charge q sits at the center of a cube of side a. If the charge is moved so that it lies at the center of one face of the cube, what is the net flux through the entire cube?

  1. q/(2ε₀)
  2. q/ε₀ Correct answer.
  3. q/(6ε₀)
  4. Zero

Explanation

The charge remains inside the closed cube, and Gauss's law states that net flux through a closed surface equals the enclosed charge divided by ε₀ regardless of where inside the charge sits. Hence the net flux is still q/ε₀.

Question 36Electric Charges, Fields, and Gauss's Law

Two point charges are separated by a fixed distance. Which change would double the magnitude of the electric force between them?

  1. Doubling one charge while leaving the other unchanged Correct answer.
  2. Doubling one charge and halving the other
  3. Halving both charges
  4. Doubling the separation between the charges

Explanation

Coulomb's law gives F = k|q₁q₂|/r², so the force is proportional to each charge. Doubling just one charge doubles the product and hence doubles the force.

Question 37Electric Charges, Fields, and Gauss's Law

A neutral solid conducting sphere is placed in a uniform external electric field and allowed to reach electrostatic equilibrium. Which statement best describes the resulting field inside the sphere's material?

  1. The field inside equals the external field, since the sphere is neutral.
  2. The field inside is the external field reduced by a factor of two.
  3. The field inside points opposite the external field with equal magnitude.
  4. The field inside is zero, and the external field lines terminate on induced surface charge. Correct answer.

Explanation

In electrostatic equilibrium the field inside a conductor must vanish, otherwise free electrons would keep moving. The external field is canceled inside by the field of induced surface charges, whose distribution produces zero net interior field.

Question 38Electric Charges, Fields, and Gauss's Law

A point charge +Q is located at the center of a neutral conducting spherical shell with inner radius a and outer radius b. What is the electric field magnitude at a point in the shell's material, a < r < b?

  1. kQ/r²
  2. kQ/a²
  3. kQ/b²
  4. Zero Correct answer.

Explanation

The point lies within conducting material in electrostatic equilibrium, where the electric field must be zero. A charge −Q is induced on the inner surface and +Q on the outer surface, and their fields cancel the field of the central charge inside the metal.

Question 39Electric Charges, Fields, and Gauss's Law

A point charge of +3.0 μC is located at the center of a spherical Gaussian surface of radius 0.20 m. If the radius of the Gaussian surface is doubled to 0.40 m while the charge remains at the center, what happens to the net electric flux through the surface?

  1. It doubles.
  2. It quadruples.
  3. It is halved.
  4. It remains the same. Correct answer.

Explanation

Gauss's law states that the net electric flux through a closed surface depends only on the enclosed charge, not on the size or shape of the surface. Since the enclosed charge is unchanged, the flux remains the same.

Question 40Electric Charges, Fields, and Gauss's Law

A very long line of charge has a uniform linear charge density λ. Which expression gives the magnitude of the electric field at a perpendicular distance r from the line, where r is much smaller than the length of the line?

  1. λ / (4πε₀ r²)
  2. λ / (2πε₀ r) Correct answer.
  3. λ / (2πε₀ r²)
  4. λ r / (2ε₀)

Explanation

For an infinite line charge, applying Gauss's law to a cylindrical surface of radius r and length L gives E(2πrL) = λL/ε₀, so E = λ/(2πε₀r). The field falls off as 1/r, not 1/r².

Question 41Electric Charges, Fields, and Gauss's Law

A point charge +Q is located at the center of a neutral conducting spherical shell with inner radius a and outer radius b. The shell is isolated. Which statement correctly describes the electric field in the region a < r < b?

  1. The field is zero because the net charge on the shell is zero.
  2. The field points radially outward and has magnitude Q/(4πε₀r²).
  3. The field is zero because the shell is a conductor. Correct answer.
  4. The field points radially inward and has magnitude Q/(4πε₀r²).

Explanation

In electrostatics, the electric field inside the material of a conductor is zero. The induced charges on the inner and outer surfaces arrange themselves to cancel the field from the point charge within the conducting material.

Question 42Electric Potential

A positive point charge is fixed at the origin. Which statement describes the electric potential V at a point a distance r away, taking V = 0 at infinity?

  1. V = kQ/r and it is negative
  2. V = 0 because the field is radial
  3. V = kQ/r² and it is positive
  4. V = kQ/r and it is positive Correct answer.

Explanation

The potential of a point charge is V = kQ/r, a scalar that falls off as 1/r. For a positive charge the potential is positive everywhere at finite distance.

Question 43Electric Potential

The electric potential at a point in space is 300 V relative to infinity. How much work must an external agent do to bring a +2.0 µC charge from infinity to that point at constant speed?

  1. 1.5×10⁻⁴ J
  2. 6.0×10⁻⁴ J Correct answer.
  3. 6.0×10⁻⁶ J
  4. 1.5×10⁻⁹ J

Explanation

Work done by an external agent at constant speed equals qV. With q = 2.0×10⁻⁶ C and V = 300 V, W = 6.0×10⁻⁴ J.

Question 44Electric Potential

The electric potential due to a set of point charges is best described as which kind of quantity?

  1. A vector directed along the net field
  2. A scalar found by summing magnitudes only
  3. A vector found by component addition
  4. A scalar found by algebraic summation Correct answer.

Explanation

Electric potential is a scalar, so contributions from different charges add algebraically with their signs. No vector components are involved.

Question 45Electric Potential

The electric field is zero everywhere inside a solid conducting sphere in electrostatic equilibrium. What is true of the potential inside the sphere?

  1. It is constant and equal to its surface value Correct answer.
  2. It decreases as 1/r toward the center
  3. It increases linearly toward the center
  4. It is zero at the center

Explanation

With E = 0 inside a conductor, the potential cannot change with position, so it equals the surface value throughout the interior.

Question 46Electric Potential

Two parallel plates are separated by 4.0 mm and carry a potential difference of 120 V. What is the magnitude of the uniform electric field between them?

  1. 4.8×10⁻¹ V/m
  2. 3.0×10⁷ V/m
  3. 3.0×10⁴ V/m Correct answer.
  4. 4.8×10² V/m

Explanation

For a uniform field, E = V/d. With V = 120 V and d = 4.0×10⁻³ m, E = 3.0×10⁴ V/m.

Question 47Electric Potential

A charge moves along an equipotential surface. What is the work done by the electric field on the charge?

  1. Equal to qV for the surface
  2. Positive if the charge is positive
  3. Zero Correct answer.
  4. Negative if the charge is negative

Explanation

Work by the field equals minus the change in potential energy, and potential energy does not change along an equipotential surface. Hence the work is zero regardless of the charge's sign.

Question 48Electric Potential

The electric potential energy of a system of two point charges q₁ and q₂ separated by distance r is given by which expression?

  1. k(q₁+q₂)/r
  2. kq₁q₂/r²
  3. kq₁q₂r
  4. kq₁q₂/r Correct answer.

Explanation

The potential energy of a pair of point charges is U = kq₁q₂/r, with the sign following the product of the charges.

Question 49Electric Potential

A +5.0 µC charge and a −5.0 µC charge are 0.20 m apart. What is the electric potential at the midpoint between them?

  1. 2.3×10⁵ V
  2. 4.5×10⁵ V
  3. 0 V Correct answer.
  4. 9.0×10⁵ V

Explanation

At the midpoint the two charges are equidistant, so their potentials kQ/r and −kQ/r cancel exactly. The scalar sum is zero.

Question 50Electric Potential

A proton is accelerated from rest through a potential difference of magnitude 500 V. What is its kinetic energy in electron-volts?

  1. 1.6×10⁻¹⁹ eV
  2. 8.0×10⁻¹⁷ eV
  3. 500 J
  4. 500 eV Correct answer.

Explanation

By definition, a particle of charge magnitude e moving through 500 V gains 500 eV of kinetic energy. The electron-volt is the energy unit that makes this direct.