301. A charge \(+1.5\,\mu\text{C}\) is moved along a path perpendicular to a uniform electric field of magnitude \(800\,\text{V m}^{-1}\) for \(0.25\,\text{m}\). What is the work done by the electric field?
ⓐ. \(+3.0\times10^{-4}\,\text{J}\)
ⓑ. \(-3.0\times10^{-4}\,\text{J}\)
ⓒ. \(0\)
ⓓ. \(+1.2\times10^{-3}\,\text{J}\)
Correct Answer: \(0\)
Explanation: \(\textbf{Given:}\) \(q=+1.5\,\mu\text{C}\), \(E=800\,\text{V m}^{-1}\), and displacement \(d=0.25\,\text{m}\).
\(\textbf{Direction condition:}\) The displacement is perpendicular to \(\vec{E}\).
\(\textbf{Work relation:}\) \(W_{\text{field}}=q\vec{E}\cdot\vec{d}\).
\(\textbf{Dot product:}\) \(\vec{E}\cdot\vec{d}=Ed\cos90^\circ\).
\(\textbf{Trigonometric value:}\) \(\cos90^\circ=0\).
\(\textbf{Result:}\) \(W_{\text{field}}=qEd(0)=0\).
\(\textbf{Potential meaning:}\) The displacement is along an equipotential direction, so \(\Delta V=0\).
\(\textbf{Final result:}\) The work done by the electric field is \(0\).
302. The potential in a region decreases most rapidly along the negative \(y\)-direction. What is the direction of the electric field?
ⓐ. Negative \(y\)-direction
ⓑ. Positive \(y\)-direction
ⓒ. Positive \(x\)-direction
ⓓ. Perpendicular to the \(xy\)-plane
Correct Answer: Negative \(y\)-direction
Explanation: Electric field points in the direction of the greatest decrease of electric potential. If the potential decreases most rapidly along the negative \(y\)-direction, then the electric field must point along that same direction. This follows from \(\vec{E}=-\nabla V\). The negative sign means the field is opposite to the direction of increasing potential. It is not decided by coordinate labels alone, but by how \(V\) changes in space. A positive test charge released from rest would tend to accelerate in the direction of \(\vec{E}\). Therefore, the field points along the negative \(y\)-direction.
303. In a uniform electric field, two equipotential planes differ by \(24\,\text{V}\). If the field magnitude is \(600\,\text{V m}^{-1}\), what is the separation between the planes?
ⓐ. \(0.010\,\text{m}\)
ⓑ. \(0.020\,\text{m}\)
ⓒ. \(0.030\,\text{m}\)
ⓓ. \(0.040\,\text{m}\)
Correct Answer: \(0.040\,\text{m}\)
Explanation: \(\textbf{Given:}\) Potential difference magnitude is \(|\Delta V|=24\,\text{V}\), and \(E=600\,\text{V m}^{-1}\).
\(\textbf{Required:}\) Separation \(d\) between equipotential planes.
\(\textbf{Uniform field relation:}\) \(E=\frac{|\Delta V|}{d}\).
\(\textbf{Solving for \(d\):}\) \(d=\frac{|\Delta V|}{E}\).
\(\textbf{Substitution:}\) \(d=\frac{24}{600}\,\text{m}\).
\(\textbf{Calculation:}\) \(d=0.040\,\text{m}\).
\(\textbf{Unit check:}\) \(\frac{\text{V}}{\text{V m}^{-1}}=\text{m}\).
\(\textbf{Final result:}\) The separation is \(0.040\,\text{m}\).
304. Which situation must have zero electric field in a region?
ⓐ. The potential is zero at one point only
ⓑ. The potential is constant throughout the region
ⓒ. The potential is positive throughout the region
ⓓ. The potential changes uniformly with distance
Correct Answer: The potential is constant throughout the region
Explanation: Electric field depends on the spatial change of potential, not on the absolute value of potential at a single point. If the potential is the same at every point of a region, then there is no potential gradient. In one dimension, this is expressed as \(E=-\frac{dV}{dx}\), and the derivative of a constant potential is zero. In three dimensions, the equivalent statement is \(\vec{E}=-\nabla V\). A potential that is positive everywhere can still produce a field if it changes with position. A potential that changes uniformly gives a constant non-zero electric field. Therefore, only constant potential throughout the region guarantees zero electric field.
305. In a uniform electric field, a displacement \(d\) is made at an angle \(120^\circ\) with the field direction. Which relation gives the potential difference between final and initial points?
ⓐ. \(V_f-V_i=+\frac{Ed}{2}\)
ⓑ. \(V_f-V_i=-\frac{Ed}{2}\)
ⓒ. \(V_f-V_i=+Ed\)
ⓓ. \(V_f-V_i=0\)
Correct Answer: \(V_f-V_i=+\frac{Ed}{2}\)
Explanation: \(\textbf{General relation:}\) In a uniform electric field, \(V_f-V_i=-Ed\cos\theta\).
\(\textbf{Given angle:}\) \(\theta=120^\circ\).
\(\textbf{Trigonometric value:}\) \(\cos120^\circ=-\frac{1}{2}\).
\(\textbf{Substitution:}\) \(V_f-V_i=-Ed\left(-\frac{1}{2}\right)\).
\(\textbf{Sign simplification:}\) The product of two negative signs gives a positive result.
\(\textbf{Final expression:}\) \(V_f-V_i=+\frac{Ed}{2}\).
\(\textbf{Physical meaning:}\) The displacement has a component opposite to the electric field, so the potential increases.
\(\textbf{Final result:}\) The potential difference is \(+\frac{Ed}{2}\).
306. A uniform electric field is \(400\,\text{V m}^{-1}\). A displacement of \(0.30\,\text{m}\) is made at \(120^\circ\) with the field direction. What is the potential difference \(V_f-V_i\)?
ⓐ. \(-120\,\text{V}\)
ⓑ. \(+60\,\text{V}\)
ⓒ. \(-60\,\text{V}\)
ⓓ. \(+120\,\text{V}\)
Correct Answer: \(+60\,\text{V}\)
Explanation: \(\textbf{Given:}\) \(E=400\,\text{V m}^{-1}\), \(d=0.30\,\text{m}\), and \(\theta=120^\circ\).
\(\textbf{Required:}\) \(V_f-V_i\).
\(\textbf{Formula:}\) \(V_f-V_i=-Ed\cos\theta\).
\(\textbf{Angle value:}\) \(\cos120^\circ=-\frac{1}{2}\).
\(\textbf{Substitution:}\) \(V_f-V_i=-(400)(0.30)\left(-\frac{1}{2}\right)\,\text{V}\).
\(\textbf{Intermediate simplification:}\) \((400)(0.30)=120\).
\(\textbf{Final simplification:}\) \(V_f-V_i=+60\,\text{V}\).
\(\textbf{Unit check:}\) \(\text{V m}^{-1}\times\text{m}=\text{V}\).
\(\textbf{Final result:}\) The potential difference is \(+60\,\text{V}\).
307. For a point charge \(+Q\), two spherical equipotential surfaces have radii \(r\) and \(3r\). Which statement about the potential difference between them is correct?
ⓐ. The outer surface has three times the potential of the inner surface
ⓑ. The outer surface has the same potential as the inner surface
ⓒ. The outer surface has one-third the potential of the inner surface
ⓓ. The outer surface has one-ninth the potential of the inner surface
Correct Answer: The outer surface has one-third the potential of the inner surface
Explanation: \(\textbf{Point-charge potential:}\) \(V=\frac{kQ}{r}\).
\(\textbf{Inner surface:}\) At radius \(r\), \(V_1=\frac{kQ}{r}\).
\(\textbf{Outer surface:}\) At radius \(3r\), \(V_2=\frac{kQ}{3r}\).
\(\textbf{Comparison:}\) \(V_2=\frac{1}{3}\frac{kQ}{r}\).
\(\textbf{Using \(V_1\):}\) \(V_2=\frac{V_1}{3}\).
\(\textbf{Meaning:}\) Potential decreases with distance from a positive point charge.
\(\textbf{Field contrast:}\) The electric field would become \(\frac{1}{9}\) of its initial value, but potential becomes \(\frac{1}{3}\).
\(\textbf{Final result:}\) The outer surface has one-third the potential of the inner surface.
308. Which statement is correct about the spacing of equipotential surfaces around an isolated point charge if equal potential drops are considered?
ⓐ. The surfaces are equally spaced at all distances
ⓑ. The surfaces become closer together as distance increases
ⓒ. The surfaces exist only on the equatorial plane
ⓓ. The surfaces become farther apart as distance increases
Correct Answer: The surfaces become farther apart as distance increases
Explanation: For a point charge, \(V=\frac{kQ}{r}\). Near the charge, potential changes rapidly with small changes in \(r\). Far from the charge, the same change in \(r\) produces a smaller change in potential. Therefore, to get the same potential drop at larger distances, a larger radial separation is needed. This means equal-potential-difference surfaces are closer near the charge and farther apart away from it. The electric field is stronger where equipotential surfaces are closer. The surfaces are spherical, not planar or equatorial only. Hence the spacing increases with distance from the point charge.
309. The potential varies with position as \(V=6x^2-4x\), where \(V\) is in \(\text{V}\) and \(x\) is in \(\text{m}\). What is \(E_x\) at \(x=1.0\,\text{m}\)?
ⓐ. \(-12\,\text{V m}^{-1}\)
ⓑ. \(+8\,\text{V m}^{-1}\)
ⓒ. \(-8\,\text{V m}^{-1}\)
ⓓ. \(+12\,\text{V m}^{-1}\)
Correct Answer: \(-8\,\text{V m}^{-1}\)
Explanation: \(\textbf{Given potential:}\) \(V=6x^2-4x\).
\(\textbf{Required:}\) \(E_x\) at \(x=1.0\,\text{m}\).
\(\textbf{Relation:}\) \(E_x=-\frac{dV}{dx}\).
\(\textbf{Differentiate:}\) \(\frac{dV}{dx}=12x-4\).
\(\textbf{At \(x=1.0\,\text{m}\):}\) \(\frac{dV}{dx}=12(1)-4=8\,\text{V m}^{-1}\).
\(\textbf{Apply negative sign:}\) \(E_x=-8\,\text{V m}^{-1}\).
\(\textbf{Direction meaning:}\) The field component points along the negative \(x\)-direction.
\(\textbf{Final result:}\) \(E_x=-8\,\text{V m}^{-1}\).
310. If the electric field in a region is zero, what can be concluded about the electric potential throughout that connected region?
ⓐ. It must be negative everywhere
ⓑ. It must be constant everywhere
ⓒ. It must be zero everywhere
ⓓ. It must increase linearly with distance
Correct Answer: It must be constant everywhere
Explanation: Electric field is related to the negative gradient of electric potential. In one dimension, \(E=-\frac{dV}{dx}\). If the electric field is zero everywhere in a connected region, then the spatial derivative of potential is zero throughout that region. A zero derivative means the potential does not change from point to point. Therefore, the potential is constant throughout that region. The constant value need not be zero; it depends on the chosen reference. A region can have \(V=50\,\text{V}\) everywhere and still have zero electric field. Hence zero electric field means constant potential, not necessarily zero potential.
311. Two points in a uniform electric field are separated by \(0.10\,\text{m}\) along the field direction. The potential difference is \(25\,\text{V}\) in magnitude. What separation would give a potential difference of \(75\,\text{V}\) in the same field?
ⓐ. \(0.20\,\text{m}\)
ⓑ. \(0.25\,\text{m}\)
ⓒ. \(0.30\,\text{m}\)
ⓓ. \(0.50\,\text{m}\)
Correct Answer: \(0.30\,\text{m}\)
Explanation: \(\textbf{Known separation:}\) \(d_1=0.10\,\text{m}\).
\(\textbf{Known potential difference:}\) \(|\Delta V_1|=25\,\text{V}\).
\(\textbf{Uniform field idea:}\) In a uniform field, \(|\Delta V|=Ed\) along the field direction.
\(\textbf{Field magnitude:}\) \(E=\frac{25}{0.10}=250\,\text{V m}^{-1}\).
\(\textbf{Required potential difference:}\) \(|\Delta V_2|=75\,\text{V}\).
\(\textbf{Required distance:}\) \(d_2=\frac{|\Delta V_2|}{E}\).
\(\textbf{Substitution:}\) \(d_2=\frac{75}{250}\,\text{m}\).
\(\textbf{Final result:}\) \(d_2=0.30\,\text{m}\).
312. A charge \(+2.0\,\mu\text{C}\) moves in a uniform field through a displacement \(0.50\,\text{m}\) making \(60^\circ\) with the field. If \(E=100\,\text{V m}^{-1}\), what is the work done by the electric field?
ⓐ. \(1.0\times10^{-4}\,\text{J}\)
ⓑ. \(5.0\times10^{-5}\,\text{J}\)
ⓒ. \(-5.0\times10^{-5}\,\text{J}\)
ⓓ. \(-1.0\times10^{-4}\,\text{J}\)
Correct Answer: \(5.0\times10^{-5}\,\text{J}\)
Explanation: \(\textbf{Given:}\) \(q=+2.0\,\mu\text{C}=2.0\times10^{-6}\,\text{C}\), \(E=100\,\text{V m}^{-1}\), \(d=0.50\,\text{m}\), and \(\theta=60^\circ\).
\(\textbf{Required:}\) Work done by the electric field.
\(\textbf{Work relation:}\) \(W_{\text{field}}=qEd\cos\theta\).
\(\textbf{Angle value:}\) \(\cos60^\circ=\frac{1}{2}\).
\(\textbf{Substitution:}\) \(W_{\text{field}}=(2.0\times10^{-6})(100)(0.50)\left(\frac{1}{2}\right)\,\text{J}\).
\(\textbf{Intermediate simplification:}\) \(100\times0.50\times\frac{1}{2}=25\).
\(\textbf{Calculation:}\) \(W_{\text{field}}=(2.0\times10^{-6})(25)=5.0\times10^{-5}\,\text{J}\).
\(\textbf{Final result:}\) The work done by the electric field is \(5.0\times10^{-5}\,\text{J}\).
313. Which statement about equipotential surfaces is always true in electrostatics?
ⓐ. Two equipotential surfaces can intersect at any angle
ⓑ. Surfaces at different potentials cannot intersect
ⓒ. Equipotential surfaces exist only for uniform electric fields
ⓓ. Equipotential surfaces always have spherical shape
Correct Answer: Surfaces at different potentials cannot intersect
Explanation: An equipotential surface is a surface on which the potential has a definite constant value. If two equipotential surfaces with different potentials intersected at a point, that same point would have two different potential values. A single point in an electrostatic field cannot simultaneously have two different scalar potentials. Therefore, equipotential surfaces with different potential values cannot intersect. Their shapes depend on the charge distribution. For a point charge they are spheres, while for a uniform field they are planes. Thus non-intersection is a general property, but spherical shape is not.
314. A positive point charge has equipotential surfaces of \(300\,\text{V}\), \(200\,\text{V}\), and \(100\,\text{V}\). Which surface is closest to the charge?
ⓐ. \(300\,\text{V}\)
ⓑ. \(200\,\text{V}\)
ⓒ. \(100\,\text{V}\)
ⓓ. All are equally distant
Correct Answer: \(300\,\text{V}\)
Explanation: \(\textbf{Point-charge potential:}\) For a positive point charge, \(V=\frac{kQ}{r}\).
\(\textbf{Distance dependence:}\) Since \(V\propto\frac{1}{r}\), larger potential corresponds to smaller distance.
\(\textbf{Given surfaces:}\) The potentials are \(300\,\text{V}\), \(200\,\text{V}\), and \(100\,\text{V}\).
\(\textbf{Largest potential:}\) \(300\,\text{V}\) is the largest value.
\(\textbf{Closest surface:}\) The surface with the largest potential is closest to the positive charge.
\(\textbf{Physical meaning:}\) Potential decreases as distance from a positive point charge increases.
\(\textbf{Final result:}\) The \(300\,\text{V}\) equipotential surface is closest to the charge.
315. A negative point charge has equipotential surfaces of \(-300\,\text{V}\), \(-200\,\text{V}\), and \(-100\,\text{V}\). Which surface is closest to the charge?
ⓐ. \(-100\,\text{V}\)
ⓑ. \(-200\,\text{V}\)
ⓒ. \(-300\,\text{V}\)
ⓓ. All are at the same radius
Correct Answer: \(-300\,\text{V}\)
Explanation: \(\textbf{Potential due to negative charge:}\) For a negative point charge, \(V=\frac{kq}{r}\), where \(q<0\).
\(\textbf{Magnitude dependence:}\) The magnitude \(|V|\) is larger when \(r\) is smaller.
\(\textbf{Given surfaces:}\) The potentials are \(-300\,\text{V}\), \(-200\,\text{V}\), and \(-100\,\text{V}\).
\(\textbf{Largest magnitude:}\) \(-300\,\text{V}\) has the greatest magnitude among the three.
\(\textbf{Closest surface:}\) The greatest magnitude corresponds to the smallest radius.
\(\textbf{Sign caution:}\) For negative values, \(-300\,\text{V}\) is numerically smaller than \(-100\,\text{V}\), but it is closer because its magnitude is larger.
\(\textbf{Final result:}\) The \(-300\,\text{V}\) surface is closest to the negative charge.
316. The potential in a region is \(V=20+5y\), where \(V\) is in \(\text{V}\) and \(y\) is in \(\text{m}\). What is the direction of the electric field?
ⓐ. Along \(+y\)
ⓑ. Along \(-y\)
ⓒ. Along \(+x\)
ⓓ. Along \(-x\)
Correct Answer: Along \(-y\)
Explanation: \(\textbf{Given potential:}\) \(V=20+5y\).
\(\textbf{Field-potential relation:}\) \(E_y=-\frac{dV}{dy}\).
\(\textbf{Derivative:}\) \(\frac{dV}{dy}=+5\,\text{V m}^{-1}\).
\(\textbf{Electric field component:}\) \(E_y=-5\,\text{V m}^{-1}\).
\(\textbf{Direction meaning:}\) A negative \(E_y\) means the electric field points along \(-y\).
\(\textbf{Physical check:}\) Potential increases with \(+y\), so the field points toward decreasing potential, which is \(-y\).
\(\textbf{Final result:}\) The electric field is along \(-y\).
317. A charge moves from \(A\) to \(B\) along a path where \(\vec{E}\cdot d\vec{l}=0\) at every small step. Which conclusion follows?
ⓐ. \(V_B=V_A\)
ⓑ. \(V_B>V_A\)
ⓒ. \(V_B<V_A\)
ⓓ. \(V_B=-V_A\)
Correct Answer: \(V_B=V_A\)
Explanation: \(\textbf{Given condition:}\) Along every small step, \(\vec{E}\cdot d\vec{l}=0\).
\(\textbf{Potential relation:}\) The small potential change is \(dV=-\vec{E}\cdot d\vec{l}\).
\(\textbf{Substitution:}\) Since \(\vec{E}\cdot d\vec{l}=0\), \(dV=0\) for every small displacement.
\(\textbf{Path result:}\) Adding all small changes along the path gives total potential change \(0\).
\(\textbf{Conclusion:}\) \(V_B-V_A=0\).
\(\textbf{Interpretation:}\) The path lies along an equipotential direction.
\(\textbf{Final result:}\) \(V_B=V_A\).
318. The potential at equally spaced points along a straight line is \(100\,\text{V}\), \(80\,\text{V}\), \(60\,\text{V}\), and \(40\,\text{V}\). The separation between adjacent points is \(0.10\,\text{m}\). What can be inferred?
ⓐ. The field is non-uniform and increasing
ⓑ. The field is uniform, \(200\,\text{V m}^{-1}\)
ⓒ. The field is zero because the potentials are positive
ⓓ. The field is uniform with magnitude \(20\,\text{V m}^{-1}\)
Correct Answer: The field is uniform, \(200\,\text{V m}^{-1}\)
Explanation: \(\textbf{Potential sequence:}\) The potential drops by \(20\,\text{V}\) between each pair of adjacent points.
\(\textbf{Spacing:}\) The separation between adjacent points is \(0.10\,\text{m}\).
\(\textbf{Uniformity check:}\) Equal potential drops over equal distances indicate a constant potential gradient.
\(\textbf{Field magnitude:}\) \(E=\frac{|\Delta V|}{\Delta x}\).
\(\textbf{Substitution:}\) \(E=\frac{20}{0.10}\,\text{V m}^{-1}\).
\(\textbf{Calculation:}\) \(E=200\,\text{V m}^{-1}\).
\(\textbf{Direction note:}\) The field points from higher potential toward lower potential along the line.
\(\textbf{Final result:}\) The field is uniform with magnitude \(200\,\text{V m}^{-1}\).
319. Which statement correctly compares moving a charge along and across equipotential surfaces?
ⓐ. Zero work along an equipotential; possible non-zero work between different potentials
ⓑ. Maximum work along an equipotential; zero work between different equipotential surfaces
ⓒ. Work is zero only across different equipotential surfaces
ⓓ. Work is independent of potential difference
Correct Answer: Zero work along an equipotential; possible non-zero work between different potentials
Explanation: Along a single equipotential surface, the potential difference is zero. Therefore, \(\Delta U=q\Delta V=0\), and the work done by the electric field is zero. When a charge moves from one equipotential surface to another with a different potential, \(\Delta V\neq0\). In that case, the potential energy generally changes by \(q\Delta V\). The electric field does work equal to \(-q\Delta V\). The amount of work depends on the charge and the potential difference, not simply on the geometrical path length. Thus motion along an equipotential requires no electrostatic work, while motion across equipotentials generally involves work.
320. In electrostatic equilibrium, what is the electric field inside the material of a conductor?
ⓐ. It is directed along the surface
ⓑ. It is proportional to the conductor volume
ⓒ. It is maximum at the centre
ⓓ. It is zero
Correct Answer: It is zero
Explanation: In electrostatic equilibrium, free charges inside a conductor are no longer moving. If an electric field existed inside the conducting material, the free charges would experience a force \(q\vec{E}\) and would continue to move. That would contradict the condition of electrostatic equilibrium. Therefore, charges redistribute themselves until the internal electric field becomes zero. This is why the potential throughout the conducting material becomes constant. The field may exist just outside the conductor surface, especially near charged regions. But inside the conducting material itself, the electrostatic field is zero.
